9.2.5 Speed and Acceleration, PT3 Focus Practice

Question 11:
Susan drives at an average speed of 105 km/h from town F to town G.
The journey takes 3 hours.

Susan takes 30 minutes longer for her return journey from town G to town F. Calculate the average speed, in km/h, for the return journey.

Solution:
$$ Distance from F to G = 105 km/h×3 hours =315 km Average speed for return journey = distance travelled time taken = 315 km 3 1 2  hours =90 km/h

Question 12:
Diagram below shows a trailer travelling from a factory to location P and location P to location Q. The trailer departs at 8.00 a.m.




(a) Based on the Table, calculate the total mass, in tonne, of the trailer and its load.

(b) The trailer arrived at location P at 10.00 a.m. and it stopped for 1½ hours to unload half of the concrete pipes. The trailer then continued its journey to location Q two times faster than its earlier speed. State the time, the trailer reached at location Q.


Solution:
(a)
$\begin{array}{l}\text{Mass of concrete pipe on the trailer}\\ =500\text{ kg}\times 8\\ =4000\text{ kg}\\ =\frac{4000}{1000}\\ =4\text{ tonnes}\\ \\ \text{Total mass of the trailer and its load}\\ =1.5+4.0\\ =5.5\text{ tonnes}\end{array}$

(b)
$\begin{array}{l}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \\ \text{Time taken to travel from the Factory to Location }P\\ =10.00\text{ a}\text{.m}\text{.}-8.00\text{ a}\text{.m}\text{.}\\ =2\text{ hours}\\ \\ \text{Speed of the trailer}\\ =\frac{80}{2}\\ =40\text{ km/h}\\ \\ \text{It stopped for }1\frac{1}{2}\text{ hours to unload half of the concrete pipes}\text{.}\\ \\ \text{Time taken to continue its journey to Location }Q\\ =\frac{\text{Distance}}{\text{Speed}}\\ =\frac{200}{40\times 2}\\ =2\frac{1}{2}\text{ hours}\\ \\ \text{The time the trailer reached at Location }Q\\ =1400\text{ hours or }2.00\text{ p}\text{.m}\text{.}\end{array}$