Motion Worksheet-11

Motion Worksheet-11

Study the speed-time graph of a body given here and answer the following question:

Calculate the retardation of the body.

A. 1 m/s² B. -1 m/s² C. 0 m/s² D. 1.5 m/s²

Study the speed-time graph of a body given here and answer the following questions :

Find out distance travelled by the body from A to B.

A. 36 m B. 1 m C. 15 m D. 60 m

A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time :

Speed (m/s) : 5 10 15 20 25 30

Time (s) : 0 10 20 30 40 50

The acceleration of the car.

A. –0.5 m/s² B. 1.5 m/s² C. –1.5 m/s² D. 0.5 m/s²

A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time :

Speed (m/s) : 5 10 15 20 25 30

Time (s) : 0 10 20 30 40 50

The distance travelled by the car in 50 seconds.

A. 875 m B. 750 m C. 625 m D. 1750 m

What type of motion is represented by each one of the following graph?

A. Uniform deceleration,constant speed,uniform retardation, non uniform retardarion

B. constant speed, constant velocity, uniform retardation, non uniform retardarion

C. uniform retardation, constant speed, Uniform acceleration, non uniform retardarion

D. Uniform acceleration, constant speed, uniform retardation, non uniform retardarion

Which graph represents the case of :

A cricket ball thrown vertically upwards and returning to the hands of the thrower?

A. A B. B

C. C D. None of these

Which graph represents the case of :

A trolley deceleration to a constant speed and then acceleration uniformly?

A. A B. B

C. C D. None of these

Answer:

Explanation: The graph line BC represents retardation. So, the slope of speed-time graph BC will be equal to the retardation of the body. So,

Retardation = slope of line BC = (BE/EC)

Now, in the graph given to us, we find that BE = 6 m/s and EC = 16 – 10 = 6 seconds. So, putting these values in the above relation, we get :

Retardation = = 1 m/s²

Explanation: We have studied that in a speed-time graph, the distance travelled by the body is equal to the area enclosed between the speed-time graph and the time-axis. Thus,

Distance travelled from A to B = Area under the line AB and the time axis

= Area of rectangle DABE

= DA × DE

Now, from the given graph (figure 40), we find that DA = 6 m/s and DE = 10 – 4 = 6 s. therefore,

Distance travelled from A to B = 6 × 6 = 36 m

Explanation: The speed-time graph obtained from the given readings is shown here. Note that in this case, when the time is 0, then the speed is not 0. the body has an initial speed of 5 m/s which is represented by point A.

Acceleration = slope of speed-time graph

= slope of line AF = (FG/AG)

Therefore, FG = 30 – 5 = 25 m/s

Again, at point G, the value of time is 50 seconds whereas that at point A is 0 second.

Thus, AG = 50 – 0 = 50 s

Now, putting these values of FG and AG in the above relation, we get :

Acceleration = = 0.5 m/s²

The distance travelled by the car in 50 seconds is equal to the area under the speed-time curve AF. That is, the distance travelled is equal to the area of the figure OAFH. But the figure OAFH is a trapezium. So,

Distance travelled = Area of trapezium OAFH

The two parallel sides are OA and HF whereas the height is OH. Therefore,

Distance travelled =

= 875 m