Graphical Representation of Motion - Free MCQ Practice Test with solutions,

Because velocity is the slope of the x-t graph.(dx/dt=V) and by looking at the graph it can be seen that upto time T₀ slope is constant so velocity will be constant and after time T0 slope is zero ( line is parallel to X axis) so the velocity will be zero.

The graph given is a velocity-time graph for the first 4 seconds of motion. To find the distance covered during this time interval, we need to calculate the area under the velocity-time graph.

The graph shows a straight line from (0, 0) to (4, 15), which forms a right-angled triangle with:

• Base = 4 seconds (time)
• Height = 15 m/s (velocity)

Distance covered = Area under the curve = Area of the triangle

Area of triangle = (1/2) × base × height

= (1/2) × 4 × 15

= 2 × 15 = 30 meters

Therefore, the distance covered in the first 4 seconds is 30 meters.

We have given in the information that the person is in his office for two hours hence his position must not vary for two hours, that is the horizontal part in the graph which is only observed in the graph of option c.

From the given graph we get that the area of the given graph is total distance covered that is 
=  ½ x 10 x 12
=  60m
And total time taken is 10 sec
Thus average speed is  60m / 10sec
 = 6 m/s

This can be verified graphically

ObtaIning instantaneous acceleration from graph

Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called "speed", being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s⁻¹).

Given:

Let two balls A and B have mass m_A, m_B respectively, and their initial velocities are u_A and u_B. After the collision, they will move with the same velocity, v_o. 

Given that mass of both balls are same.

So  m_A = m_B = m

u_A = V,  u_B = 0

From the Concept of Momentum Conservation:

m_Au_A + m_Bu_B = (m+m)v_o

mV = 2mv_o

v_o = V/2

Both the balls after impact will move with velocity v/2.

Option B is correct.

Coincidence of two curves at the same point means the two objects have the same value of the quantity plotted on the vertical axis; that quantity here is velocity, so their velocities are equal at that instant.

Acceleration at a point on a velocity-time curve equals the slope of the curve at that point, expressed as a = dv/dt.

If the slopes of the two curves at the point of intersection are equal, the accelerations are equal; if the slopes are different, the accelerations are different. Therefore, only velocity is certainly the same; acceleration may or may not be the same depending on the slopes.

The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. The boundary line is dashed for > and < and solid for ≤ and ≥.

Analyze the graphs:

1. Graph 1:

   • The slopes of P and Q are different.
   • Therefore, the velocities are different, and the relative velocity is not zero.

2. Graph 2:

   • Both P and Q have parallel lines, meaning the slopes are equal.
   • Equal slopes indicate equal velocities, and the relative velocity is zero.
   • This graph satisfies the condition.

3. Graph 3:

   • The slopes of P and Q are different.
   • Hence, the velocities are different, and the relative velocity is not zero.

4. Graph 4:

   • The slopes of P and Q are different.
   • Again, the velocities are different, and the relative velocity is not zero.