Graphs of Motion - NEET Physics Questions
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Graphs of Motion

Question 1:

The displacement-time graph of two moving particles A and B respectively make angles of 30° and 45° with the x-axis respectively. The ratio of their velocities is

1. 1 : √3
2. 1 : 2
3. 1 : 1
4. √3 : 2
View Answer

Slope of displacement time graph represents velocity.

VA= tan 30 0
VB= tan 45 0

So, VA/VB=tan 30 0/tan 45 0= 1/√3

Question 2:

A body is travelling in a straight line with a uniformly increasing speed. Which one of the plot represents the change in distance (s) travelled with time (t)?

1.
2.
3.
4.
View Answer

\[ s= \frac{1}{2}at^{2}\]

As s is proportional to square of t. s-t graph is a parabola.

Question 3:

A particle is thrown upwards, then correct v-t graph will be

1.
2.
3.
4.
View Answer

Slope of v-t graph represent acceleration. when object is thrown upward its acceleration is -g for the entire journey. So, slope is -g.

Question 4:

A body starts from rest and moves with uniform acceleration. Which of the following graphs represents it motion?

1.
2.
3.
4.
View Answer

\[ v = u + at ; as, u = 0 ; v = at\]

As, v is proportional to t ; graph of v-t is a straight line.

Question 5:

If the figure below represents a parabola, identify the physical quantities representing Y and X for constant acceleration

1. X = time, Y = velocity
2. X = velocity, Y = time
3. X = time, Y = displacement
4. X = time, Y = acceleration
View Answer

\[ S= \frac{1}{2}at^{2}\]

As, S is proportional to t graph is a parabola

Question 6:

A particle is thrown vertically upwards with a velocity ν. It returns to the ground in time T. Which of the following graphs correctly represents the motion?

1.
2.
3.
4.
View Answer

Graph of velocity time graph represents acceleration. During the entire journey acceleration is -g. so slope is negative.

Question 7:

A car moving with a speed of 50 km h–¹ can be stopped by brakes after atleast 6m. If the same car is moving at a speed of 100 km h–¹ the minimum stopping distance is :

1. 6 m
2. 12 m
3. 18 m
4. 24 m
View Answer

The stopping distance is proportional to the square of the speed:
\[
\frac{s_2}{s_1} = \left( \frac{v_2}{v_1} \right)^2
\]

Substituting the values:
\[
\frac{s_2}{6} = \left( \frac{100}{50} \right)^2 = 2^2 = 4
\]
\[
s_2 = 6 \times 4 = 24 \, \text{m}
\]

Thus, the minimum stopping distance at 100 km/h is 24 meters.

Question 8:

The displacement-time graph of a moving particle is as shown in the figure. The instantaneous velocity of the particle is negative at the point :

1. C
2. D
3. E
4. F
View Answer

Slope of x-t graph represents velocity. Slope is negative at point E so, velocity is negative for E.