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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
Correct answer is option 'A'. Can you explain this answer?
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
|
Hansa Sharma answered |
Given: x2 = at2 + b
Differentiating w.r.t t on both sides, we get
Again differentiating w.r.t on both sides, we get
Differentiating w.r.t t on both sides, we get
Again differentiating w.r.t on both sides, we get
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
Correct answer is option 'D'. Can you explain this answer?
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
When its velocity is 1 m s−1, its acceleration isa)
2 m s-2
b)
7 m s-2
c)
1 m s-2
d)
0.5 m s-2
|
Mira Joshi answered |
Given: 
Squaring both sides, we get, v2=49+x
Differentiating both sides w.r.t. t, we get,
Differentiating both sides w.r.t. t, we get,
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
Correct answer is option 'A'. Can you explain this answer?
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
|
Raghav Bansal answered |
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 A to B,
For the motion of car from B to C
S = 9km = 9000m
For the motion of car from C to D,
For the motion of car from C to D,
Total time taken = t1 + t2 + t3 ;
= 50s + 900s + 100s = 1050s
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.
Correct answer is option 'D'. Can you explain this answer?
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.
|
Ananya Das answered |
For uniform motion with zero acceleration, v-t graph is a straight line parallel to the time axis.
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
Correct answer is option 'A'. Can you explain this answer?
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
|
|
Mira Joshi answered |
As x ∝ t3
velocity, v ∝ 3t2
Acceleration, a ∝ 6t
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
Correct answer is option 'B'. Can you explain this answer?
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
|
Dev Patel answered |
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
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
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
Correct answer is option 'A'. Can you explain this answer?
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
|
|
Anjali Sharma answered |
The slope of the tangent drown on a velocity-time graph at any instant of time is equal to the instantaneous acceleration.
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)
Correct answer is option 'D'. Can you explain this answer?
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
will bea)
b)
c)
d)
|
|
Mira Joshi answered |
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
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
Correct answer is option 'A'. Can you explain this answer?
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
|
|
Mira Joshi answered |
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.
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.
The area under acceleration-time graph gives- a)initial velocity
- b)final velocity
- c)change in velocity
- d)distance travelled
Correct answer is option 'C'. Can you explain this answer?
The area under acceleration-time graph gives
a)
initial velocity
b)
final velocity
c)
change in velocity
d)
distance travelled
|
|
Ajay Yadav answered |
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.
Position-time graph for motion with zero acceleration is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Position-time graph for motion with zero acceleration is
a)
b)
c)
d)
|
|
Raghav Bansal answered |
For zero acceleration, the position-time graph is a straight line.
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)
Correct answer is option 'D'. Can you explain this answer?
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)
|
Priya Menon answered |
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.
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)
Correct answer is option 'A'. Can you explain this answer?
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)
|
Gaurav Kumar answered |
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.
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
Correct answer is option 'D'. Can you explain this answer?
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
|
Tanish Mahajan answered |
Let's discuss each graph one by one.
A) we can see that when the graph gets +ve, at the axis the graph returns back to the axis indicating 0 displacement.
B) at the peak of 2nd graph we can see that slope is 0 so velocity and acceleration at that point would be 0.
C) 3rd graph has -ve slope this velocity is also -ve
D) 4th graph has +ve slope so velocity is also +ve
A) we can see that when the graph gets +ve, at the axis the graph returns back to the axis indicating 0 displacement.
B) at the peak of 2nd graph we can see that slope is 0 so velocity and acceleration at that point would be 0.
C) 3rd graph has -ve slope this velocity is also -ve
D) 4th graph has +ve slope so velocity is also +ve
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