Motion in a Straight Line

NEET Physics · 98 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s^-2, the velocity with which it hits the ground is

A 10.0 m/s

B 20.0 m/s

C 40.0 m/s

D 5.0 m/s

Correct Answer

Option B

Solution

v = sqrt(2gh) = sqrt(2*10*20) = 20 m/s

A particle moves a distance x in time t according to equation x = (t+5)^-1. The acceleration of particle is proportional to

A (velocity)^(3/2)

B (distance)^2

C (distance)^(-2)

D (velocity)^(2/3)

Correct Answer

Option A

Solution

v = -1/(t+5)^2, a = 2/(t+5)^3. a = -2*v^(3/2), so a proportional to (velocity)^(3/2)

A ball is dropped from high rise at t=0. After 6s another ball is thrown downward with speed v. They meet at t=18s. What is v? (g=10 m/s^2)

A 75 m/s

B 55 m/s

C 40 m/s

D 60 m/s

Correct Answer

Option A

Solution

1st ball distance in 18s = (1/2)*10*18^2 = 1620 m. For 2nd ball in 12s: 1620 = v*12 + (1/2)*10*144. So v=75 m/s

A bus moves at 10 m/s. A scooterist wishes to overtake in 100s. Bus is 1km away. What speed should scooterist use?

A 40 m/s

B 25 m/s

C 10 m/s

D 20 m/s

Correct Answer

Option D

Solution

1000/(vs-10) = 100, so vs = 20 m/s

A particle starts from rest under constant force. Distance in first 10s is S1 and first 20s is S2. Then:

A S2 = 3S1

B S2 = 4S1

C S2 = S1

D S2 = 2S1

Correct Answer

Option B

Solution

S = (1/2)*a*t^2. S1/S2 = (10/20)^2 = 1/4, so S2 = 4S1

A particle moving along x-axis has acceleration f = f0*(1 - t/T). At t=0 velocity is zero. Velocity when f=0:

A (1/2)*f0*T^2

B f0*T^2

C (1/2)*f0*T

D f0*T

Correct Answer

Option C

Solution

f=0 when t=T. vx = integral from 0 to T of f0*(1-t/T)dt = f0*[t - t^2/(2T)]_0^T = f0*(T - T/2) = (1/2)*f0*T

A car moves from X to Y with speed vu and returns with vd. Average speed for round trip:

A sqrt(vu*vd)

B vd*vu/(vd+vu)

C (vu+vd)/2

D 2*vd*vu/(vd+vu)

Correct Answer

Option D

Solution

Average speed = 2S / (S/vu + S/vd) = 2*vu*vd/(vu+vd)

Position x = 9t^2 - t^3. Position when it achieves maximum speed along +x:

A 54 m

B 81 m

C 24 m

D 32 m

Correct Answer

Option A

Solution

v = 18t - 3t^2. dv/dt = 18-6t = 0, t=3s. x = 9*9 - 27 = 54 m

x = 40 + 12t - t^3. Distance travelled before coming to rest:

A 16 m

B 24 m

C 40 m

D 56 m

Correct Answer

Option A

Solution

v = 12 - 3t^2 = 0 at t=2s. s = integral_0^2 (12-3t^2)dt = [12t-t^3]_0^2 = 24-8 = 16 m

Q10

A car runs at constant speed on circular track r=100m, taking 62.8s per lap. Average velocity and speed respectively:

A 10 m/s, 0

B 0, 0

C 0, 10 m/s

D 10 m/s, 10 m/s

Correct Answer

Option C

Solution

Distance = 2*pi*100 = 628m. Speed = 628/62.8 = 10 m/s. Net displacement per loop = 0, average velocity = 0.

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