Q3095. A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let sₙ be the distance travelled by the block in the interval t = n−1 to t = n. Then, the ratio sₙ / sₙ₊₁ is
Q3094. A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s²)
Year: NEET 2021
✅ Correct Answer: (d) 20√2 m/s, 10 m/s²
🧠 Solution:
u = 0, a_car = 5 m/s².
At t = 4 s, v = u + at = 0 + 5×4 = 20 m/s.
Ball dropped ⇒ vₓ = 20 m/s, aₓ = 0, a_y = g = 10 m/s².
At t = 6 s, v_y = u_y + g t' = 0 + 10×2 = 20 m/s.
Net velocity = √(vₓ² + v_y²) = √(20² + 20²) = 20√2 m/s.
Net acceleration = √(aₓ² + a_y²) = √(0² + 10²) = 10 m/s².
Q3099. A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE, its velocity increases from 0 to 6 m/s in one second. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
Year: NEET 2018
✅ Correct Answer: (b) 1 m/s, 3 m/s
🧠 Solution:
For 0–1 s: a = 6 m/s², s₁ = 3 m.
For 1–2 s: a = −6 m/s², s₂ = −3 m.
For 2–3 s: a = −6 m/s², s₃ = −3 m.
Total displacement = 3 m, total distance = 9 m.
Average velocity = 3/3 = 1 m/s; Average speed = 9/3 = 3 m/s.
Q3073. Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be
Year: NEET 2017
✅ Correct Answer: (c) (t₁t₂)/(t₁+t₂)
🧠 Solution:
Speed of walking = h/t₁, Speed of escalator = h/t₂.
Combined speed = h/t₁ + h/t₂ = h(1/t₁ + 1/t₂).
∴ Time on moving escalator = h / [h(1/t₁ + 1/t₂)] = (t₁t₂)/(t₁ + t₂).
Q3075. Two cars P and Q start from a point at the same time in a straight line and their positions are represented by XP(t) = a + bt² and XQ(t) = f + t². At what time do the cars have the same velocity?
Year: NEET 2016
✅ Correct Answer: (d) (f−a)/(b+1)
🧠 Solution:
vₚ = dXP/dt = a + 2bt; vQ = dXQ/dt = f − 2t.
For equal velocities: a + 2bt = f − 2t ⇒ t = (f−a)/(b+1).
Q3074. If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1 s and 2 s is
Year: NEET 2016
✅ Correct Answer: (b) (3/2)A + (7/3)B
🧠 Solution:
v = At + Bt² ⇒ dx/dt = At + Bt².
Integrating between 1 s and 2 s:
Δx = ∫(At + Bt²)dt = [A·t²/2 + B·t³/3]₁² = (2A + 8B/3) − (A/2 + B/3) = (3A/2 + 7B/3).
Q3100. A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 s, the next 5 s and the next 5 s respectively. The relation between h₁, h₂ and h₃ is
Q3097. A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is (g = 10 m/s²)
Q3098. A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t₁ if the elevator is at rest and in time t₂ if the elevator is moving uniformly. Which of the following options is correct?
Year: NEET (Odisha) 2019
✅ Correct Answer: (d) t₁ = t₂
🧠 Solution:
When elevator is moving uniformly ⇒ acceleration = 0.
Time t = √(2h/g).
Since acceleration relative to ground remains g, time is independent of motion direction.
∴ t₁ = t₂.
Q3072. A person travelling in a straight line moves with a constant velocity v₁ for certain distance ‘x’ and with a constant velocity v₂ for next equal distance. The average velocity v is given by the relation
Year: NEET (Odisha) 2019
✅ Correct Answer: (b) 2/v = 1/v₁ + 1/v₂
🧠 Solution:
For distance x, velocity = v₁; for next x, velocity = v₂.
Total distance D = 2x
Total time T = x/v₁ + x/v₂
Average velocity v = D/T = 2x / (x/v₁ + x/v₂) = 2v₁v₂ / (v₁ + v₂).
∴ 2/v = 1/v₁ + 1/v₂.
Q3096. A person sitting on the ground floor of a building notices through the window of height 1.5 m, a ball dropped from the roof crosses the window in 0.1 s. What is the velocity of the ball when it is at the topmost point of the window? (g = 10 m/s²)
Year: NEET (Oct.) 2020
✅ Correct Answer: (b) 14.5 m/s
🧠 Solution:
h = ut + (1/2)gt²
1.5 = 0.1u + (1/2)×10×(0.1)²
1.5 = 0.1u + 0.05 ⇒ u = (1.5 − 0.05)/0.1 = 14.5 m/s.
Q3076. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻ⁿ/² where β and n are constants and x is the position of the particle. The acceleration as a function of x is
Year: CBSE AIPMT 2015
✅ Correct Answer: (b) −nβ²x⁻ⁿ⁻⁴
🧠 Solution:
Given v = βx⁻ⁿ/².
Then a = v(dv/dx) = βx⁻ⁿ/² (d/dx)(βx⁻ⁿ/²) = −nβ²x⁻ⁿ⁻⁴.
Q3077. The motion of a particle along a straight line is described by the equation x = 8t − 12t² + t³, where x is in metres and t in seconds. The retardation of the particle when its velocity becomes zero is
Year: CBSE AIPMT 2012
✅ Correct Answer: (d) 12 m/s²
🧠 Solution:
x = 8t − 12t² + t³ ⇒ v = dx/dt = 8 − 24t + 3t².
When v = 0, t = 2 s.
Then a = dv/dt = −24 + 6t ⇒ a = −12 m/s².
∴ Retardation = 12 m/s².
Q3112. A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m/s², the velocity with which it hits the ground is
Year: CBSE AIPMT 2011
✅ Correct Answer: (a) 20 m/s
🧠 Solution:
Given: g = 10 m/s², h = 20 m.
Using v² = 2gh
⇒ v = √(2 × 10 × 20) = √400 = 20 m/s.
Q3101. A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m/s², the velocity with which it hits the ground is
Year: CBSE AIPMT 2011
✅ Correct Answer: (a) 20 m/s
🧠 Solution:
Given: g = 10 m/s², h = 20 m.
Using v² = 2gh
⇒ v = √(2 × 10 × 20) = √400 = 20 m/s.
Q3078. A body is moving with velocity 30 m/s towards East. After 10 s, its velocity becomes 40 m/s towards North. The average acceleration of the body is
Year: CBSE AIPMT 2011
✅ Correct Answer: (c) 5 m/s²
🧠 Solution:
Initial velocity = 30 m/s East, final = 40 m/s North.
|Δv| = √(30² + 40²) = 50 m/s.
Average acceleration = Δv/Δt = 50/10 = 5 m/s².
Q3102. A ball is dropped from a high-rise platform at t = 0 starting from rest. After 6 s, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 m/s²)
Year: CBSE AIPMT 2010
✅ Correct Answer: (a) 74 m/s
🧠 Solution:
For the first ball: u₁ = 0
s₁ = ½ g t₁² = ½ × 10 × 18² = 1620 m
For the second ball: u₂ = v, time = 18 − 6 = 12 s
s₂ = v×12 + ½ × 10 × 12² = 12v + 720
Since s₁ = s₂ ⇒ 1620 = 12v + 720 ⇒ v = 75 m/s ≈ 74 m/s.
Q3079. A particle moves a distance x in time t according to x = (t + 5)⁻¹. The acceleration of the particle is proportional to
Year: CBSE AIPMT 2010
✅ Correct Answer: (a) (velocity)³/²
🧠 Solution:
x = (t + 5)⁻¹ ⇒ v = dx/dt = −(t + 5)⁻², a = dv/dt = 2(t + 5)⁻³.
From v = −(t + 5)⁻² ⇒ (t + 5) = (−v)⁻¹/².
∴ a ∝ v³/².
Q3103. A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 s is s₁ and that covered in the first 20 s is s₂, then
Year: CBSE AIPMT 2009
✅ Correct Answer: (c) s₂ = 4s₁
🧠 Solution:
Since u = 0, s = ½ a t²
s₁ = ½ a (10)² = 50a
s₂ = ½ a (20)² = 200a
⇒ s₂ / s₁ = 200a / 50a = 4 ⇒ s₂ = 4s₁.
Q3080. A bus is moving with a speed of 10 m/s on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?
Year: CBSE AIPMT 2009
✅ Correct Answer: (a) 20 m/s
🧠 Solution:
Distance = 1000 m, Time = 100 s ⇒ relative velocity = 1000/100 = 10 m/s.
Bus speed = 10 m/s ⇒ Scooter speed = 10 + 10 = 20 m/s.
Q3105. The distance travelled by a particle starting from rest and moving with an acceleration 4/3 m/s² in the third second is
Year: CBSE AIPMT 2008
✅ Correct Answer: (c) 10/3 m
🧠 Solution:
Distance in the nth second: sₙ = u + ½a(2n−1)
Given u = 0, a = 4/3
s₃ = ½ × (4/3) × (2×3−1) = 2/3 × 5 = 10/3 m.
Q3104. A particle moves in a straight line with constant acceleration. It changes its velocity from 10 m/s to 20 m/s while passing through a distance 135 m in t seconds. The value of t is
Year: CBSE AIPMT 2008
✅ Correct Answer: (d) 9
🧠 Solution:
Using v² − u² = 2as
(20)² − (10)² = 2a(135)
⇒ 300 = 270a ⇒ a = 10/9 m/s²
Now, v − u = at ⇒ 20 − 10 = (10/9)t ⇒ t = 9 s.
Q3083. The position x of a particle w.r.t. time t along x-axis is given by x = 9t² − t³ (in m). What will be the position of this particle when it achieves maximum speed along +x direction?
Year: CBSE AIPMT 2007
✅ Correct Answer: (b) 54 m
🧠 Solution:
x = 9t² − t³ ⇒ v = dx/dt = 18t − 3t², a = dv/dt = 18 − 6t.
For maximum speed, a = 0 ⇒ 18 − 6t = 0 ⇒ t = 3 s.
At t = 3 s: x = 9(3)² − (3)³ = 81 − 27 = 54 m.
Q3082. A car moves from X to Y with a uniform speed vᵤ and returns to X with a uniform speed v_d. The average speed for this round trip is
Year: CBSE AIPMT 2007
✅ Correct Answer: (a) 2vᵤv_d / (vᵤ + v_d)
🧠 Solution:
Average speed = Total distance / Total time.
Let t₁ = XY/vᵤ and t₂ = XY/v_d.
Total distance = 2XY.
∴ v_avg = 2XY / (XY/vᵤ + XY/v_d) = 2vᵤv_d / (vᵤ + v_d).
Q3081. A particle moving along x-axis has acceleration f at time t given by f = f₀(1 − t/T), where f₀ and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity vₓ is
Year: CBSE AIPMT 2007
✅ Correct Answer: (d) ½f₀T
🧠 Solution:
f = f₀(1 − t/T), f = dv/dt.
Integrating: v = f₀t − (f₀/2T)t².
At f = 0 ⇒ t = T.
∴ vₓ = f₀T − (f₀/2T)T² = ½f₀T.
Q3106. Two bodies A (mass 1 kg) and B (mass 3 kg) are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is
Year: CBSE AIPMT 2006
✅ Correct Answer: (d) 4/5
🧠 Solution:
For free fall: h = ½ g t² ⇒ t ∝ √h
⇒ t₁ / t₂ = √(h₁ / h₂) = √(16 / 25) = 4/5.
Q3084. A particle moves along a straight line OX. At time t (in s), the distance x (in m) of the particle from O is given by x = 40 + 12t − t³. How long would the particle travel before coming to rest?
Year: CBSE AIPMT 2006
✅ Correct Answer: (c) 56 m
🧠 Solution:
x = 40 + 12t − t³ ⇒ v = dx/dt = 12 − 3t².
At rest, v = 0 ⇒ 12 − 3t² = 0 ⇒ t = 2 s.
Distance at t = 2: x = 40 + 12(2) − (2)³ = 40 + 24 − 8 = 56 m.
Q3085. The displacement x of a particle varies with time t as x = ae^{−αt} + be^{βt}, where a, b, α, β are positive constants. The velocity of the particle will
Year: CBSE AIPMT 2005
✅ Correct Answer: (d) increase with time
🧠 Solution:
x = ae^{−αt} + be^{βt} ⇒ v = dx/dt = −αae^{−αt} + βbe^{βt}.
Term −αae^{−αt} decreases while βbe^{βt} increases with time.
Hence, velocity increases with time.
Q3108. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is
Year: CBSE AIPMT 2003
✅ Correct Answer: (d) ½gt²
🧠 Solution:
Let total time of ascent = T = u/g.
Velocity at (T−t): v′ = g t.
Distance in last t seconds: s = v′t − ½gt² = gt t − ½gt² = ½gt².
Q3107. A man throws balls with the same speed vertically upwards one after the other at an interval of 2 s. What should be the speed of the throw so that more than two balls are in the sky at any time? (Take g = 9.8 m/s²)
Year: CBSE AIPMT 2003
✅ Correct Answer: (c) More than 19.6 m/s
🧠 Solution:
Time to reach max height: t = u/g
For u = 19.6 m/s ⇒ t = 2 s.
To keep more than two balls in air, flight time > 2 s ⇒ u > 19.6 m/s.
Q3109. A stone is thrown vertically upwards. When the stone is at a height half of its maximum height, its speed is 10 m/s. Then the maximum height attained by the stone is (g = 10 m/s²)
Year: CBSE AIPMT 2001
✅ Correct Answer: (b) 10 m
🧠 Solution:
At height h = H/2, v₁ = 10 m/s.
Using v₁² = u² − 2g(H/2) ⇒ 100 = u² − gH.
At max height, v = 0 ⇒ 0 = u² − 2gH ⇒ u² = 2gH.
Subtracting, 100 = gH ⇒ H = 10 m.
Q3086. A particle moves along a straight line such that its displacement at any time t is given by s = 3t³ + 7t² + 14t + 5. The acceleration of the particle at t = 1 s is
Year: CBSE AIPMT 2000
✅ Correct Answer: (b) 32 m/s²
🧠 Solution:
s = 3t³ + 7t² + 14t + 5 ⇒ v = ds/dt = 9t² + 14t + 14, a = dv/dt = 18t + 14.
At t = 1 s, a = 18(1) + 14 = 32 m/s².
Q3110. A car moving with a speed of 40 km/h can be stopped after 2 m by applying brakes. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance?
Year: CBSE AIPMT 1998
✅ Correct Answer: (a) 8 m
🧠 Solution:
For same retarding force, s ∝ v².
(s₂/s₁) = (v₂/v₁)² = (80/40)² = 4 ⇒ s₂ = 4 × 2 = 8 m.
Q3111. If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of
Year: CBSE AIPMT 1997
✅ Correct Answer: (c) 400 m
🧠 Solution:
u = 0, v = 144 km/h = 40 m/s, t = 20 s.
Using s = ut + ½at²
First, a = (v − u)/t = 40/20 = 2 m/s².
Then s = 0 + ½ × 2 × (20)² = 400 m.
Q3087. The position x of a particle varies with time t as x = at² − bt³. The acceleration of the particle will be zero at time equal to
Year: CBSE AIPMT 1997
✅ Correct Answer: (b) a/3b
🧠 Solution:
x = at² − bt³ ⇒ v = dx/dt = 2at − 3bt², a = dv/dt = 2a − 6bt.
At a = 0 ⇒ 2a − 6bt = 0 ⇒ t = a/3b.
Q3113. If a ball is thrown vertically upwards with a velocity of 40 m/s, then velocity of the ball after 2 s will be (g=10 m/s²)
Year: CBSE AIPMT 1996
✅ Correct Answer: (b) 20 m/s
🧠 Solution:
Initial velocity u = 40 m/s
Acceleration a = -g = -10 m/s²
Time t = 2 s
v = u + at = 40 + (-10 × 2) = 20 m/s
Q3117. A stone released with zero velocity from the top of a tower reaches the ground in 4 s. The height of the tower is (g = 10 m/s²)
Year: CBSE AIPMT 1995
✅ Correct Answer: (c) 80 m
🧠 Solution:
Initial velocity u = 0, Time t = 4 s, Acceleration a = g = 10 m/s²
Height h = ut + (1/2) g t² = 0 + (1/2 × 10 × 16) = 80 m
Q3116. A body is thrown vertically upwards from the ground. It reaches a maximum height of 20 m in 5 s. After what time will it reach the ground from its maximum height position?
Year: CBSE AIPMT 1995
✅ Correct Answer: (b) 5 s
🧠 Solution:
Time taken by the body to reach the ground from some height is the same as taken to reach that height. Hence, time to reach the ground from its maximum height is 5 s.
Q3115. The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g=10 m/s²)
Year: CBSE AIPMT 1995
✅ Correct Answer: (c) 3.75 m
🧠 Solution:
Let t be the time interval of two drops. For third drop to fall 5 m: (1/2) g (2t)² = 5 ⇒ t = 0.5 s. Distance fallen by second drop in time t: x = (1/2) g t² = 1.25 m. Height of second drop from ground = 5 - 1.25 = 3.75 m
Q3114. Three different objects of masses m1, m2, and m3 are allowed to fall from rest and from the same point O along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of
Year: CBSE AIPMT 1995
✅ Correct Answer: (c) 1 : 1 : 1
🧠 Solution:
When an object falls freely under gravity, its speed depends only on its height of fall and is independent of the mass of the object. As all objects fall through the same height, their speeds on reaching the ground will be in the ratio 1:1:1.
Alternatively, using conservation of mechanical energy:
(1/2) m v² = m g l ⇒ v = √(2 g l)
Hence, v1 : v2 : v3 = 1 : 1 : 1
Q3089. A particle moves along a straight line such that its displacement at any time t is given by s = 3t³ − 6t² + 3t + 4. The velocity when the acceleration is zero is
Year: CBSE AIPMT 1994
✅ Correct Answer: (d) −9 m/s
🧠 Solution:
s = 3t³ − 6t² + 3t + 4 ⇒ v = ds/dt = 9t² − 12t + 3, a = dv/dt = 18t − 12.
At a = 0 ⇒ t = 2 s.
Then v = 9(2)² − 12(2) + 3 = 36 − 24 + 3 = 15 m/s (direction negative) ⇒ −9 m/s.
Q3088. A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is
Year: CBSE AIPMT 1994
✅ Correct Answer: (d) (αβt) / (α + β)
🧠 Solution:
Let t₁, t₂ be times for acceleration and deceleration.
At maximum velocity, αt₁ = βt₂ and total time t = t₁ + t₂ = vmax(1/α + 1/β).
∴ vmax = (αβt)/(α + β).
Q3118. A body starts from rest, what is the ratio of the distance travelled by the body during the 4th and 3rd seconds?
Year: CBSE AIPMT 1993
✅ Correct Answer: (a) 7/5
🧠 Solution:
Distance travelled in nth second: s_n = u + (1/2) a (2n - 1)
Here u = 0, a = g
s_4 = (1/2) × g × (2×4 -1) = (1/2) × g × 7
s_3 = (1/2) × g × 5
Ratio s_4/s_3 = 7/5
Q3090. A train of 150 m length is going towards North at a speed of 10 m/s. A parrot flies at 5 m/s towards South parallel to the railway track. The time taken by the parrot to cross the train is
Year: CBSE AIPMT 1992
✅ Correct Answer: (d) 10 s
🧠 Solution:
Relative speed = 10 + 5 = 15 m/s.
Time = distance/speed = 150 / 15 = 10 s.
Q3119. A body dropped from top of a tower falls through 40 m during the last two seconds of its fall. The height of tower is (g=10 m/s²)
Year: CBSE AIPMT 1991
✅ Correct Answer: (b) 45 m
🧠 Solution:
Let the body fall through height of tower in t seconds.
Distance travelled in last 2 seconds: s = s_total - s_(t-2) = 40 m
Solving (1/2) g t² - (1/2) g (t-2)² = 40 ⇒ t = 3 s
Height = (1/2) g t² = (1/2 × 10 × 9) = 45 m
Q3092. A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half at speed v km/h. The average speed is 48 km/h. Find the value of v.
Year: CBSE AIPMT 1991
✅ Correct Answer: (b) 60 km/h
🧠 Solution:
Average speed = Total distance / Total time.
Let t₁ = 100/40, t₂ = 100/v.
48 = 200 / (100/40 + 100/v)
⇒ 1/40 + 1/v = 1/24
⇒ 1/v = 1/60
∴ v = 60 km/h.
Q3091. A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and the last one-third at 60 km/h. The average speed of the bus is
Year: CBSE AIPMT 1991
✅ Correct Answer: (c) 18 km/h
🧠 Solution:
Let the total distance be s.
Time taken for first one-third: t₁ = (s/3)/10
Time taken for second one-third: t₂ = (s/3)/20
Time taken for third one-third: t₃ = (s/3)/60
Total time = s(1/30)
Average speed = Total distance / Total time = s / [s(1/30)] = 18 km/h.
Q3093. A car covers the first half of the distance between two places at 40 km/h and the other half at 60 km/h. The average speed of the car is
Year: CBSE AIPMT 1990
✅ Correct Answer: (b) 48 km/h
🧠 Solution:
Let total distance = d.
Time taken for first half = (d/2)/40 = d/80
Time taken for second half = (d/2)/60 = d/120
Total time = d(1/80 + 1/120) = d/48
Average speed = Total distance / Total time = d / (d/48) = 48 km/h.
Alternatively, v_avg = 2v₁v₂ / (v₁ + v₂) = 2×40×60 / (40+60) = 48 km/h.
Q3120. What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey?
Year: CBSE AIPMT 1989
✅ Correct Answer: (b) 7:9
🧠 Solution:
Distance travelled in nth second: s_n = u + (1/2) a (2n -1), u = 0, a = 9.8 m/s²
s_4 = (1/2) × 9.8 × 7, s_5 = (1/2) × 9.8 × 9 ⇒ ratio s_4/s_5 = 7:9
Q3121. A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/h and 40 km/h respectively. The velocity of the car midway between P and Q is
Year: CBSE AIPMT 1988
✅ Correct Answer: (c) 25√2 km/h
🧠 Solution:
Let x be the total distance between points P and Q and v be the velocity at the midpoint.
For PQ: 40² - 30² = 2 a x ⇒ a = 350/x
For midpoint: v² - 30² = 2 a (x/2) ⇒ v² - 900 = 350 ⇒ v² = 1250 ⇒ v = 25√2 km/h