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

Question 7:


Solution:
(a)
Distance travel by the object for the first 12 seconds

Question 8:


Solution:
(a)
$\begin{array}{l}\text{Acceleration from }t=6s\text{ to }t=12s\\ =\text{ Gradient of speed-time graph}\\ =\frac{6-6}{12-6}\\ =0{\text{ ms}}^{-2}\leftarrow \boxed{\begin{array}{l}\text{The moving object is moving}\\ \text{ at a uniform speed}\end{array}}\end{array}$

(c)

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

$\begin{array}{l}\text{(b)}\\ \text{Rate of change of speed for the first 8 seconds}\\ =\frac{10-6}{0-8}\\ =-\frac{4}{8}\\ =-\frac{1}{2}{\text{ ms}}^{-2}\end{array}$

$\begin{array}{l}\text{(c)}\\ \text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \frac{\left(\frac{1}{2}\times 12\times \left(6+u\right)\right)+\left(6\times u\right)+\left(\frac{1}{2}\times 8\times u\right)}{26}=6\\ 36+6u+6u+4u=156\\ 16u=120\\ u=7.5{\text{ ms}}^{-1}\end{array}$