Class 11Physics · MechanicsFull chapter

Motion in a Straight Line

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Position, Path Length and DisplacementAverage and Instantaneous Velocity and SpeedAverage and Instantaneous AccelerationPosition-Time and Velocity-Time GraphsKinematic Equations of Uniformly Accelerated MotionRelative Velocity in One Dimension

Position, Path Length and Displacement

Quick answer Position is measured from a chosen origin along a direction; path length is the total distance travelled (scalar), while displacement is the vector change in position, and its magnitude is always less than or equal to the path length.

To describe motion along a straight line we first fix an origin O and a positive direction on that line. The position of an object at any instant is its distance from the origin, given a sign depending on which side of the origin it lies. Position is meaningful only with respect to a chosen origin and direction.

Path length (also called distance) is the total length of the actual route travelled between two instants. It is a scalar, is always positive, and can never decrease with time.

Displacement is the change in position: Δx = x2 − x1, where x1 and x2 are the initial and final positions. For straight-line motion, displacement is a vector — it may be positive, negative or zero, and its sign shows the direction of the net change of position.

For any motion, |Δx| ≤ s (path length). The two are equal only when the object moves without ever reversing direction. If the object reverses direction, s exceeds |Δx|; if it returns exactly to its starting point, Δx = 0 even though s ≠ 0.

Worked Example

  • Given: A car starts at the origin (x = 0), moves to x = +200 m, then reverses to reach x = +50 m.
  • Formula: Path length s = sum of magnitudes of each leg; Displacement Δx = xfinal − xinitial.
  • Substitution: Leg 1: 0 → +200 m, length = 200 m. Leg 2: +200 m → +50 m, length = |50 − 200| = 150 m. Displacement: Δx = 50 − 0.
  • Result: Path length s = 200 m + 150 m = 350 m; Displacement Δx = +50 m. Note s > |Δx| because the car reversed direction.
Displacement Δx = x₂ − x₁ x₁, x₂ are initial and final positions from the origin; the sign shows direction
Displacement–path length inequality |Δx| ≤ s s = total path length; equality only for motion without reversal
Remember
  • Position needs a reference origin and a chosen positive direction.
  • Path length (distance) is a scalar, always ≥ 0, and never decreases with time.
  • Displacement Δx = x₂ − x₁ is a vector for straight-line motion; it can be positive, negative or zero.
  • |Displacement| ≤ Path length always; equality holds only for one-directional motion.
  • Displacement can be zero for non-zero path length if the object returns to its start.

Average and Instantaneous Velocity and Speed

Quick answer Average velocity is net displacement over time (a vector), average speed is total path length over time (a scalar, always ≥ average velocity's magnitude), and instantaneous velocity is the slope of the x-t graph at a point.

Average velocity over an interval is the displacement divided by the time taken: vavg = Δx/Δt = (x2 − x1)/(t2 − t1). Since displacement is a vector, average velocity is also a vector and carries a sign in one dimension.

Average speed over the same interval is total path length divided by time: average speed = s/Δt. It is a scalar, always positive, and always ≥ the magnitude of average velocity, since s ≥ |Δx|. They are equal only for one-directional motion.

Instantaneous velocity at instant t is the limit of average velocity as Δt → 0: v = limΔt→0 Δx/Δt = dx/dt — the slope of the tangent to the position-time (x-t) graph at that instant. Instantaneous speed is simply |v|, the magnitude of instantaneous velocity, and (unlike the average quantities) it always exactly equals the magnitude of instantaneous velocity.

Worked Example

  • Given: Position of a particle is x(t) = 3t² + 2t (x in metres, t in seconds). Find average velocity between t = 1 s and t = 3 s, and instantaneous velocity at t = 2 s.
  • Formula: vavg = [x(t2) − x(t1)]/(t2 − t1); instantaneous v = dx/dt.
  • Substitution: x(1) = 3(1)² + 2(1) = 5 m; x(3) = 3(3)² + 2(3) = 33 m. vavg = (33 − 5)/(3 − 1) = 28/2. Also v(t) = 6t + 2, so v(2) = 6(2) + 2.
  • Result: Average velocity = 14 m/s; Instantaneous velocity at t = 2 s = 14 m/s (equal here since t = 2 s is the midpoint of [1 s, 3 s] and v is linear in t).
Average velocity v_avg = Δx/Δt = (x₂ − x₁)/(t₂ − t₁) vector; sign shows direction of net displacement
Average speed average speed = s/Δt scalar; s = total path length; average speed ≥ |v_avg|
Instantaneous velocity v = dx/dt slope of tangent to the x-t graph at that instant
Instantaneous speed speed = |v| magnitude of instantaneous velocity
Remember
  • Average velocity = Δx/Δt (vector); average speed = total path length/Δt (scalar).
  • Average speed ≥ |average velocity|; equal only when motion is one-directional.
  • Instantaneous velocity v = dx/dt = slope of the tangent to the x-t graph.
  • Instantaneous speed always equals |instantaneous velocity|, exactly.
  • A straight x-t graph means uniform velocity; a curved one means changing velocity.

Average and Instantaneous Acceleration

Quick answer Average acceleration is the change in velocity over time; instantaneous acceleration is the slope of the v-t graph, and its sign relative to velocity tells whether the object is speeding up or slowing down.

Average acceleration over an interval is the change in velocity divided by the time taken: aavg = Δv/Δt = (v2 − v1)/(t2 − t1). Acceleration is a vector for straight-line motion; its sign shows the direction it points along the line — not simply whether the object is speeding up or slowing down.

Instantaneous acceleration at instant t is the limit of average acceleration as Δt → 0: a = limΔt→0 Δv/Δt = dv/dt = d²x/dt² — the slope of the tangent to the velocity-time (v-t) graph at that instant.

If acceleration acts in the same direction as velocity, speed increases; if it acts opposite to velocity, speed decreases (often called retardation or deceleration). The SI unit of acceleration is m s−2.

Worked Example

  • Given: A car's velocity decreases uniformly from 20 m/s to 8 m/s in 4 s while braking.
  • Formula: aavg = (v − u)/t.
  • Substitution: aavg = (8 − 20)/4 = −12/4.
  • Result: aavg = −3 m/s². The negative sign shows the acceleration is opposite to the direction of motion — the car decelerates at 3 m/s².
Average acceleration a_avg = Δv/Δt = (v₂ − v₁)/(t₂ − t₁) vector quantity; SI unit m s⁻²
Instantaneous acceleration a = dv/dt = d²x/dt² slope of tangent to the v-t graph; also the second derivative of position
Remember
  • Average acceleration = Δv/Δt; instantaneous acceleration a = dv/dt = d²x/dt².
  • Acceleration is a vector; its sign shows direction, not just speeding-up/slowing-down.
  • Acceleration in the direction of velocity increases speed; opposite direction decreases it (retardation).
  • Instantaneous acceleration is the slope of the tangent to the v-t graph.
  • SI unit of acceleration is m s⁻².

Position-Time and Velocity-Time Graphs

Quick answer The slope of an x-t graph gives velocity and the slope of a v-t graph gives acceleration; the area under a v-t graph between two instants gives displacement.

A position-time (x-t) graph plots position against time. Its slope at any instant equals the instantaneous velocity at that instant. A straight-line x-t graph means uniform velocity (constant slope); a curved x-t graph means velocity is changing, i.e., the object has acceleration.

A velocity-time (v-t) graph plots velocity against time. Its slope at any instant equals the instantaneous acceleration. Crucially, the area enclosed between the v-t graph and the time axis, between two instants, equals the displacement in that interval — area above the axis is positive, area below is negative.

For zero acceleration (uniform velocity), the v-t graph is a horizontal line and the x-t graph is an oblique straight line. For uniform (constant) acceleration, the v-t graph is an oblique straight line and the x-t graph is a parabola.

Worked Example

  • Given: The v-t graph of a particle is a straight line from u = 5 m/s at t = 0 to v = 25 m/s at t = 4 s.
  • Formula: a = slope = (v − u)/t; displacement s = area under v-t graph = ½(u + v)t (trapezium area).
  • Substitution: a = (25 − 5)/4 = 20/4. s = ½(5 + 25)(4) = ½(30)(4).
  • Result: a = 5 m/s²; s = 60 m. (Check: s = ut + ½at² = 5(4) + ½(5)(4)² = 20 + 40 = 60 m — matches.)
Slope of x-t graph slope = dx/dt = v gives instantaneous velocity
Slope of v-t graph slope = dv/dt = a gives instantaneous acceleration
Area under v-t graph Area(t₁ to t₂) = Δx displacement; area below the time axis counts as negative
Remember
  • Slope of x-t graph = instantaneous velocity.
  • Slope of v-t graph = instantaneous acceleration.
  • Area under v-t graph between t₁ and t₂ = displacement in that interval.
  • Straight x-t graph ⇒ uniform velocity; parabolic x-t graph ⇒ uniform acceleration.
  • Horizontal v-t graph ⇒ zero acceleration; oblique straight v-t graph ⇒ uniform acceleration.

Kinematic Equations of Uniformly Accelerated Motion

Quick answer For constant acceleration, v = u + at, s = ut + ½at², and v² = u² + 2as connect initial velocity, final velocity, acceleration, time and displacement; all three can be derived from the v-t graph.

When an object moves with uniform (constant) acceleration a along a straight line, three standard equations connect initial velocity u, final velocity v, acceleration a, time t and displacement s. They can be derived graphically from the v-t graph, a straight line of slope a starting at v = u.

First equation (v = u + at): The slope between (0, u) and (t, v) is a = (v − u)/t. Rearranging gives v = u + at.

Second equation (s = ut + ½at²): Displacement equals the area under the v-t graph between 0 and t. This trapezium splits into a rectangle of height u and width t (area ut) and a triangle of height at and width t (area ½·t·at = ½at²). Adding gives s = ut + ½at².

Third equation (v² = u² + 2as): Eliminating t between the first two (substitute t = (v − u)/a into s = ½(u + v)t, the trapezium's area) gives s = (v² − u²)/(2a), i.e., v² = u² + 2as.

These equations apply only while acceleration is constant in magnitude and direction; a sign convention must be fixed first, and u, v, a, s substituted with correct signs.

Worked Example

  • Given: A train starts from rest and accelerates uniformly at 2 m/s² for 10 s.
  • Formula: v = u + at; s = ut + ½at²; check with v² = u² + 2as.
  • Substitution: u = 0, a = 2 m/s², t = 10 s. v = 0 + (2)(10). s = 0(10) + ½(2)(10)².
  • Result: v = 20 m/s; s = ½(2)(100) = 100 m. Check: v² = 0 + 2(2)(100) = 400 ⇒ v = 20 m/s — consistent.
First equation of motion v = u + at
Second equation of motion s = ut + ½at²
Third equation of motion v² = u² + 2as
Distance in the n-th second sₙ = u + (a/2)(2n − 1) derived from s = ut + ½at²; n is the second number (1st, 2nd, ...)
Remember
  • v = u + at, s = ut + ½at², v² = u² + 2as hold only for constant acceleration.
  • All three equations follow from the straight-line v-t graph (slope = a; area = s).
  • A consistent sign convention for u, v, a, s is essential before substituting.
  • Distance in the n-th second, sₙ = u + (a/2)(2n − 1), is a useful extension for numericals.
  • Vertical motion under gravity is a special case with a = ±g.

Relative Velocity in One Dimension

Quick answer The velocity of one object as observed from another moving object is their velocity difference, v_AB = v_A − v_B, with a fixed sign convention along the line of motion.

When two objects A and B move along the same straight line, the velocity of A relative to B is the rate at which A's position changes as seen by an observer moving with B. With a fixed positive direction chosen for both:

vAB = vA − vB

If A and B move in the same direction, relative velocity is the difference of their (signed) velocities. If they move in opposite directions, their velocities carry opposite signs in the chosen convention, so the relative velocity works out to the sum of their magnitudes — this is why two objects moving towards each other appear to approach faster than either moves alone.

If both are accelerating, relative acceleration is similarly aAB = aA − aB, and the standard kinematic equations can be applied directly to the relative quantities (relative velocity, relative acceleration, relative displacement) to solve problems involving approach, overtaking, or objects released at different times.

Worked Example

  • Given: On parallel tracks, train A moves at 72 km/h and train B moves at 90 km/h, both in the same direction.
  • Formula: vBA = vB − vA, after converting speeds to SI units (1 km/h = 5/18 m/s).
  • Substitution: vA = 72 × 5/18 = 20 m/s; vB = 90 × 5/18 = 25 m/s. vBA = 25 − 20.
  • Result: vBA = 5 m/s. To a passenger in train A, train B appears to move forward at only 5 m/s, even though B's actual speed is 25 m/s.
Relative velocity of A w.r.t. B v_AB = v_A − v_B fix a sign convention first; opposite-direction motion effectively adds magnitudes
Relative acceleration of A w.r.t. B a_AB = a_A − a_B
Remember
  • Relative velocity of A w.r.t. B: v_AB = v_A − v_B, using one fixed sign convention.
  • Same-direction motion: relative velocity is the (signed) difference of velocities.
  • Opposite-direction motion: relative velocity magnitude equals the sum of speeds.
  • Relative acceleration a_AB = a_A − a_B; kinematic equations apply to relative quantities.
  • Relative velocity problems commonly appear as overtaking or approach/collision numericals.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

Δx = x₂ − x₁
Displacement
|Δx| ≤ s
Displacement–path length inequality
v_avg = Δx/Δt = (x₂ − x₁)/(t₂ − t₁)
Average velocity
average speed = s/Δt
Average speed
v = dx/dt
Instantaneous velocity
speed = |v|
Instantaneous speed
a_avg = Δv/Δt = (v₂ − v₁)/(t₂ − t₁)
Average acceleration
a = dv/dt = d²x/dt²
Instantaneous acceleration
slope = dx/dt = v
Slope of x-t graph
slope = dv/dt = a
Slope of v-t graph
Area(t₁ to t₂) = Δx
Area under v-t graph
v = u + at
First equation of motion
s = ut + ½at²
Second equation of motion
v² = u² + 2as
Third equation of motion
sₙ = u + (a/2)(2n − 1)
Distance in the n-th second
v_AB = v_A − v_B
Relative velocity of A w.r.t. B
a_AB = a_A − a_B
Relative acceleration of A w.r.t. B

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Position and Displacement easy

Which of the following quantities associated with the motion of a body along a straight line is a vector?

Q2 Position and Displacement medium

For a particle moving in a straight line without ever reversing its direction, the magnitude of its displacement is:

Q3 Velocity and Speed easy

A body moves 4 m towards east and then 3 m towards west, the whole journey taking 10 s. The average speed of the body is:

Q4 Velocity and Speed medium

The position of a particle moving along the x-axis is x = 2t² + 3t (x in metres, t in seconds). Its instantaneous velocity at t = 2 s is:

Q5 Graphs easy

On a velocity-time graph, the area enclosed between the graph and the time axis, over a given interval, represents:

Q6 Graphs easy

The slope of a position-time (x-t) graph at any instant gives the:

Q7 Kinematic Equations medium

A particle starts from rest and moves with a uniform acceleration of 4 m/s². The distance covered by it in the first 3 seconds is:

Q8 Kinematic Equations hard

A car moving at 20 m/s is brought to rest by uniform deceleration in 5 s. The magnitude of the deceleration and the distance travelled before stopping are respectively:

Q9 Acceleration easy

Which of the following is true for a body moving with uniform velocity along a straight line?

Q10 Graphs medium

The velocity-time graph of a particle is a straight line from (t = 0, v = 10 m/s) to (t = 5 s, v = 0). The acceleration of the particle is:

Q11 Relative Velocity medium

Two trains A and B move in the same direction along parallel tracks with speeds 20 m/s and 15 m/s respectively. The velocity of train A relative to train B is:

Q12 Relative Velocity hard

A stone is dropped from a height, and exactly 1 s later a second identical stone is dropped from the same point. As both stones fall freely under gravity, the separation between them:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a tower 25 m high. Taking g = 10 m/s² and the upward direction as positive, find (a) the maximum height above the ground reached by the ball, and (b) the time taken by the ball to reach the ground.Kinematic Equations

Take the point of projection (top of the tower) as the origin, with upward direction positive. Then u = +20 m/s, a = −g = −10 m/s², and the ground is at position x = −25 m.

(a) Maximum height above the point of projection: At the highest point, v = 0. Using v² = u² + 2as: 0 = (20)² + 2(−10)s ⇒ 0 = 400 − 20s ⇒ s = 20 m above the top of the tower. Maximum height above the ground = 25 + 20 = 45 m.

(b) Time to reach the ground: The ground is at x = −25 m. Using s = ut + ½at²: −25 = 20t − 5t² ⇒ 5t² − 20t − 25 = 0 ⇒ t² − 4t − 5 = 0 ⇒ (t − 5)(t + 1) = 0. So t = 5 s (the root t = −1 s is not physical).

The ball reaches a maximum height of 45 m above the ground and takes 5 s to hit the ground.

2 Explain, with the help of a suitable example each, the distinction between (a) magnitude of displacement and total path length covered by an object, and (b) average speed and magnitude of average velocity of an object.Position and Displacement

(a) Magnitude of displacement vs path length: Displacement is the straight-line change in position (a vector), while path length is the actual length of the route travelled (a scalar). For example, if an object moves from x = 0 to x = +10 m and then returns to x = +4 m, the path length is 10 m + 6 m = 16 m, but the magnitude of displacement is only |4 − 0| = 4 m. The magnitude of displacement can never exceed the path length.

(b) Average speed vs magnitude of average velocity: Average speed is total path length divided by total time (always positive), while average velocity is total displacement divided by total time. For example, an athlete who runs once around a 400 m circular track and returns to the starting point in 80 s has an average speed of 400/80 = 5 m/s, but since the net displacement is zero, the magnitude of the average velocity is 0 m/s. In general, average speed ≥ magnitude of average velocity.

3 A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?Kinematic Equations

Given: u = 126 km/h, v = 0, s = 200 m. Convert to SI units: u = 126 × 5/18 = 35 m/s.

Retardation: Using v² = u² + 2as: 0 = (35)² + 2a(200) ⇒ 0 = 1225 + 400a ⇒ a = −1225/400 = −3.0625 m/s². The retardation (magnitude of deceleration) is approximately 3.06 m/s².

Time to stop: Using v = u + at: 0 = 35 + (−3.0625)t ⇒ t = 35/3.0625 ≈ 11.4 s.

4 On a two-lane road, car A is travelling at 36 km/h. Two cars, B and C, each moving at 54 km/h, approach A from opposite directions — B from behind (same direction as A) and C from the front (opposite direction). At an instant when AB = AC = 1 km, B decides to overtake A before C meets A. What minimum acceleration of car B is required to avoid an accident?Relative Velocity

Given: vA = 36 km/h = 10 m/s (positive direction), vB = 54 km/h = 15 m/s (same direction as A), vC = 54 km/h = 15 m/s but opposite direction, so vC = −15 m/s. Initial separations AB = AC = 1000 m.

Time available before C meets A: Relative velocity of C w.r.t. A = vC − vA = −15 − 10 = −25 m/s, i.e., C approaches A at 25 m/s. Time to cover the 1000 m gap: t = 1000/25 = 40 s.

Relative motion of B w.r.t. A: Initial relative velocity of B w.r.t. A = vB − vA = 15 − 10 = 5 m/s. For B to safely overtake A, B must cover the 1000 m gap (relative to A) within t = 40 s, with relative acceleration a (equal to B's own acceleration, since A moves at constant velocity).

Using s = ut + ½at² for the relative motion: 1000 = (5)(40) + ½a(40)² ⇒ 1000 = 200 + 800a ⇒ 800a = 800 ⇒ a = 1 m/s².

The minimum acceleration required for car B is 1 m/s².

5 State, with reasons and an example for each, whether the following situations are possible: (a) an object with constant acceleration but with zero velocity; (b) an object moving in a certain direction with an acceleration in the opposite direction.Acceleration

(a) Possible. At the highest point of a ball thrown vertically upward, its velocity is instantaneously zero, but the acceleration due to gravity (g, directed downward) continues to act on it throughout, including at that instant. So an object can have zero velocity and non-zero, constant acceleration simultaneously.

(b) Possible. When a ball is thrown vertically upward, its velocity is directed upward throughout the ascent, but the acceleration due to gravity acts downward (opposite to the velocity) the entire time — this is exactly why the ball slows down, momentarily stops, and then falls back.

6 A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief's car?Relative Velocity

Given: Speed of police van, vvan = 30 km/h = 30 × 5/18 = 25/3 m/s. Speed of thief's car, vthief = 192 km/h = 192 × 5/18 = 160/3 m/s. Muzzle speed of bullet relative to the van = 150 m/s.

Speed of the bullet relative to the ground: Since the bullet is fired in the same direction as the van's motion, vbullet = 150 + 25/3 = 475/3 m/s.

Speed of the bullet relative to the thief's car: vrel = vbullet − vthief = 475/3 − 160/3 = 315/3 = 105 m/s.

The bullet hits the thief's car at a relative speed of 105 m/s.

Previous-year board questions 4

Q1 The displacement of a body is proportional to the cube of the time elapsed, i.e., x ∝ t³. What is the nature of the acceleration of the body? CBSE 2020 1 mark

Let x = kt³, where k is a constant. Then velocity v = dx/dt = 3kt², and acceleration a = dv/dt = 6kt.

Since a ∝ t, the acceleration is not constant — it increases linearly with time (the motion is non-uniformly accelerated).

Q2 A car travels the first half of the distance between two places with a speed of 30 km/h and the second half with a speed of 50 km/h. Calculate the average speed of the car for the whole journey. CBSE 2022 2 marks

Let the total distance be 2d, so each half is d. Time for first half, t1 = d/30. Time for second half, t2 = d/50.

Total time = t1 + t2 = d/30 + d/50 = d(5 + 3)/150 = 8d/150 = 4d/75.

Average speed = total distance ÷ total time = 2d ÷ (4d/75) = 150/4 = 37.5 km/h.

Q3 Derive the equation of motion v² = u² + 2as graphically, where the symbols have their usual meanings. CBSE 2019 3 marks

Consider a velocity-time graph that is a straight line, starting at velocity u at t = 0 and reaching velocity v at time t, since the acceleration a is constant.

The slope of this line gives the acceleration: a = (v − u)/t, so t = (v − u)/a ... (i)

The displacement s equals the area under the v-t graph, a trapezium with parallel sides u and v and width t: s = ½(u + v)t ... (ii)

Substituting (i) into (ii): s = ½(u + v) × (v − u)/a = (v² − u²)/(2a).

Therefore, 2as = v² − u², i.e., v² = u² + 2as, which is the required equation of motion.

Q4 A stone is dropped from the top of a tower 100 m high. Simultaneously, another stone is thrown up from the foot of the tower with a velocity of 25 m/s. Find when and where the two stones meet. (Take g = 10 m/s²) CBSE 2023 5 marks

Take the foot of the tower as the origin, upward direction positive. The top of the tower is at x = 100 m.

Stone A (dropped from the top): x = 100 m, uA = 0, a = −10 m/s². Position: xA(t) = 100 + 0·t + ½(−10)t² = 100 − 5t².

Stone B (thrown up from the foot): x = 0, uB = +25 m/s, a = −10 m/s². Position: xB(t) = 25t − 5t².

Meeting condition: xA(t) = xB(t) ⇒ 100 − 5t² = 25t − 5t² ⇒ 100 = 25t ⇒ t = 4 s.

Position where they meet: xB(4) = 25(4) − 5(4)² = 100 − 80 = 20 m. (Check: xA(4) = 100 − 5(16) = 20 m ✓)

The two stones meet after 4 s, at a height of 20 m above the ground (80 m below the top of the tower).

Stuck? Ask about this chapter. Type your question in your own words — it searches straight through Motion in a Straight Line for the answer. The Ask AI button (bottom-right) stays scoped to this chapter while you're here, too.

Part of Priodemy for School

Interactive Maths & Science — free with every school on Priodemy EduSuite. Explore more chapters and labs on the Priodemy for School hub.

Ask AI