Test: Acceleration (NCERT) - NEET MCQ

# Test: Acceleration (NCERT) - NEET MCQ

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## 15 Questions MCQ Test NCERTs at Fingertips: Textbooks, Tests & Solutions - Test: Acceleration (NCERT)

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Test: Acceleration (NCERT) - Question 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

Detailed Solution for Test: Acceleration (NCERT) - Question 1

As x ∝ t3

velocity, v  ∝ 3t2

Acceleration, a ∝ 6t

Test: Acceleration (NCERT) - Question 2

### Match the Column I with Column II.

Detailed Solution for Test: Acceleration (NCERT) - Question 2

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Test: Acceleration (NCERT) - Question 3

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

Detailed Solution for Test: Acceleration (NCERT) - Question 3

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

Test: Acceleration (NCERT) - Question 4

The area under acceleration-time graph gives

Detailed Solution for Test: Acceleration (NCERT) - Question 4

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.

Test: Acceleration (NCERT) - Question 5

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

Detailed Solution for Test: Acceleration (NCERT) - Question 5

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

Test: Acceleration (NCERT) - Question 6

Match the column I with column II

Test: Acceleration (NCERT) - Question 7

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

Detailed Solution for Test: Acceleration (NCERT) - Question 7

Given:

Squaring both sides, we get, v2=49+x
Differentiating both sides w.r.t. t, we get,

Test: Acceleration (NCERT) - Question 8

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

Detailed Solution for Test: Acceleration (NCERT) - Question 8

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

Again differentiating w.r.t on both sides, we get

Test: Acceleration (NCERT) - Question 9

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

Detailed Solution for Test: Acceleration (NCERT) - Question 9

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 = t+ t+ t3 ;

= 50s + 900s + 100s = 1050s

Test: Acceleration (NCERT) - Question 10

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

Detailed Solution for Test: Acceleration (NCERT) - Question 10

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

Test: Acceleration (NCERT) - Question 11

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

Detailed Solution for Test: Acceleration (NCERT) - Question 11

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

Test: Acceleration (NCERT) - Question 12

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

Detailed Solution for Test: Acceleration (NCERT) - Question 12

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.

Test: Acceleration (NCERT) - Question 13

Position-time graph for motion with zero acceleration is

Detailed Solution for Test: Acceleration (NCERT) - Question 13

For zero acceleration, the position-time graph is a straight line.

Test: Acceleration (NCERT) - Question 14

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?

Detailed Solution for Test: Acceleration (NCERT) - Question 14

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.

Test: Acceleration (NCERT) - Question 15

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.

Detailed Solution for Test: Acceleration (NCERT) - Question 15

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.

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