Motion in a Straight Line Test

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Motion in a Straight Line Test - 1

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Question 1
4 / -1

The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is

A
increasing with time

B
decreasing with time

C
constant but not zero

D
zero

Solution

As x ∝ t³
velocity, v ∝ 3t²
Acceleration, a ∝ 6t
Question 2
4 / -1

Match the Column I with Column II.

A
A−p, B−q, C−r, D−s

B
A−r, B−s, C−p, D−q

C
A−s, B−p, C−q, D−r

D
A−q, B−r, C−s, D−p

Solution
Question 3
4 / -1

The slope of the tangent drawn on velocity-time graph at any instant of time is equal to the instantaneous

A
acceleration

B
velocity

C
impulse

D
momentum

Solution

The slope of the tangent drown on a velocity-time graph at any instant of time is equal to the instantaneous acceleration.
Question 4
4 / -1

The area under acceleration-time graph gives

A
initial velocity

B
final velocity

C
change in velocity

D
distance travelled

Solution

The vertical axis will represent the acceleration of the object. The slope of the acceleration graph will represent a quantity called the jerk. This jerk is the rate of change of the acceleration. The area under this acceleration graph represents the change in velocity. Also, this area under the acceleration-time graph for some time interval will be the change in velocity during that time interval. Multiplying this acceleration by the time interval will be equivalent to finding the area under the curve.
Question 5
4 / -1

Which of the following statements is not correct regarding the motion of a particle in a straight line?

A
x-t graph is a parabola, if motion is uniformly accelerated.

B
v-t is a straight line inclined to the time-axis, if motion is uniformly accelerated.

C
x-t graph is a straight line inclined to the time axis if motion is uniform and acceleration is zero.

D
v-t graph is a parabola if motion is uniform and acceleration is zero.

Solution

For uniform motion with zero acceleration, v-t graph is a straight line parallel to the time axis.
Question 6
4 / -1

Match the column I with column II

A
A-p, B-q, C-s, D-r

B
A-q, B-p,C-r,D-s

C
A-s, B-r, C-q, D-p

D
A-r, B-q, C-s, D-p
Question 7
4 / -1

A particle moving along a straight line has a velocity v m s⁻¹, when it cleared a distance of x m. These two are connected by the relation When its velocity is 1 m s⁻¹, its acceleration is

A
2 m s^-2

B
7 m s^-2

C
1 m s^-2

D
0.5 m s^-2

Solution

Given:
Squaring both sides, we get, v²=49+x
Differentiating both sides w.r.t. t, we get,
Question 8
4 / -1

A particle moves rectilinearly. its displacement x at time t is given by x² = t² + 1. Its acceleration at time t is proportional to

A
1/x³

B
1/x - 1/x²

C
- t/x²

D
1/x - t²/x³

Solution

Given: x²= at²+ b
Differentiating w.r.t t on both sides, we get

Again differentiating w.r.t on both sides, we get
Question 9
4 / -1

A car starts from rest, attains a velocity of 36 km h⁻¹ with an acceleration of 0.2 m s⁻², travels 9 km with this uniform velocity and then comes to halt with a uniform deceleration of 0.1 m s⁻². The total time of travel of the car is

A
1050 s

B
1000 s

C
950 s

D
900 s

Solution

Let the car be accelerated from A to B. It moves with uniform velocity from B to C and then moves with uniform declaration from C to D.
For the motion of car from A to B,

For the motion of car from B to C
S = 9km = 9000m

For the motion of car from C to D,
Total time taken = t₁+ t₂+ t₃ ;
= 50s + 900s + 100s = 1050s
Question 10
4 / -1

A particle moving with uniform acceleration has average velocities v₁, v₂ and v₃ over the successive intervals of time t₁, t₂ and t₃ respectively. The value of will be

A

B

C

D

Solution

Let u be initial velocity and a be uniform acceleration.

Average velocities in the intervals from 0 to t₁, t₁ to t₂ and t₂ to t₃ are

Subtract (i) from (ii), we get

Subtract (ii) from (iii), we get

Divide (iv) by (v), we get
Question 11
4 / -1

A particle starts from point A moves along a straight line path with an acceleration given by a = p − qx, where p,q are constants and x is distance from point A. The particle stops at point B. The maximum velocity of the particle is

A
p/q

B
p/√q

C
q/p

D
√q/p

Solution

Given: a = p − qx
At maximum velocity, a = 0
⇒ 0 = p - qx or x = p/q
∴ Velocity is maximum at x = p/q
since Acceleration

vdv = (p − qx)dx
Integrating both sides of the above equation, we get
Question 12
4 / -1

For the one dimensional motion, described by x = t - sint

A
x(t) > 0 for all t > 0

B
v(t) > 0 for all t > 0

C
a(t) > 0 for all t > 0

D
v(t) lies between 0 and 2

Solution

x = t - sint;
v = dx/dt = 1 - cost;
a = dv/dt = sint
∴ x(t) > 0 for all values of t > 0 and v(t) can be zero for one value of t. a(t) can zero for one value of t.
Question 13
4 / -1

Position-time graph for motion with zero acceleration is

A

B

C

D

Solution

For zero acceleration, the position-time graph is a straight line.
Question 14
4 / -1

The velocity-displacement graph of a particle is as shown in the figure. Which of the following graphs correctly represents the variation of acceleration with displacement?

A

B

C

D

Solution

For the given velocity-displacement graph, intercept = v₀ and slope = -v₀/x₀
Thus, the equation of given line of velocity-displacement graph is
It is a straight line with positive slope and a negative intercept.
The variation of a with x is as shown in the above figure.
Question 15
4 / -1

The velocity-time graph of a particle in one-dimensional motion is shown in Fig.
Which of the following formula are correct for describing the motion of the particle over the time-interval t₁ to t₂.

A

B

C

D

Solution

in the given interval the slop of v-t graph (i.e acceleration) is neither constant nor uniform. Therefore relations (a), (b) and (c) are incorrect but (d) is correct.