6.3 Gradient and Area Under a Graph, SPM Paper 2 (Long Questions)


Question 7:

The diagram above shows the speed-time graph of a moving object for 15 seconds. Find
(a)    the speed of the object at t = 9s.
(b)   the distance travel by the object for the first 12 seconds.

Solution:

(a)
The speed of the object at t = 9s is 6 ms-1

(b)
Distance travel by the object for the first 12 seconds
= Area under the speed-time graph
= Area of triangle
= ½ × 12 × 8
= 48 m




Question 8:

The diagram above shows the speed-time graph of a moving object for 15 seconds.
(a) Find the time interval when the object moves with constant speed.
(b) Find the acceleration from t = 6 s to t = 12 s.
(c) State the time when the object is stationary.

Solution:

(a)
Time interval when the object moves with constant speed
= 12 – 6
= 6 s

(b)
Acceleration from t=6s to t=12s = Gradient of speed-time graph = 66 126 =0  ms 2 The moving object is moving  at a uniform speed

(c)

Time when the object is stationary is at 15s.



Question 9:
Diagram below shows the speed-time graph for the movement of an object for a period of 34 seconds.


(a) State the duration of time, in seconds, for which the object moves with uniform speed.
(b) Calculate the rate of change of speed, in ms-2, of the object for the first 8 seconds.
(c) Calculate the value of u, if the average of speed of the object for the last 26 seconds is 6 ms-1.

Solution:
(a) Duration of time = 26s – 20s = 6s

(b) Rate of change of speed for the first 8 seconds = 106 08 = 4 8 = 1 2  ms 2

(c) Speed= Distance Time ( 1 2 ×12×( 6+u ) )+( 6×u )+( 1 2 ×8×u ) 26 =6                                   36+6u+6u+4u=156                                                        16u=120                                                            u=7.5  ms 1



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