
STUCK ON HOMEWORK?
ASK FOR HELP FROM OUR KIASUPARENTS.COM COMMUNITY!
Question

Answer

a) Retardation of 1.4 m/s2 during the first 10 seconds means that the car travels 1.4 m/s slower every second. So for 10 seconds, his speed would have decreased a total of (1.4 x 10) m/s.
u (initial speed) → 24 + (1.4 x 10)
= 24 + 14
= 36 m/s
b) Average Speed = Total Distance / Total Time
The Total Distance is the area under the speed/time graph.
Distance for first 10 sec = Area of trapezium on the left
= 1/2 x (24 + 36) x 10
= 1/2 x 60 x 10
= 300 m
Distance for next 30 sec = Area of bottom right triangle
= 1/2 x 30 x 24
= 360 m
Total distance travelled → 300 + 360
= 660 m
Average Speed = 660 m / 40 s
= 16.5 m/s
c) See attached picture
d) Lorry travelled at constant speed, so for the entire 40 sec duration, it would have travelled:
25 m/s x 40 s = 1,000 m
We know that the car travelled a total of 660 m during the 40 sec, so the lorry would be (1,000 – 660) m ahead.
1000 – 660 = 340 m

a) Retardation of 1.4 m/s2 during the first 10 seconds means that the car travels 1.4 m/s slower every second. So for 10 seconds, his speed would have decreased a total of (1.4 x 10) m/s.
u (initial speed) → 24 + (1.4 x 10)
= 24 + 14
= 36 m/s
b) Average Speed = Total Distance / Total Time
The Total Distance is the area under the speed/time graph.
Distance for first 10 sec = Area of trapezium on the left
= 1/2 x (24 + 36) x 10
= 1/2 x 60 x 10
= 300 m
Distance for next 30 sec = Area of bottom right triangle
= 1/2 x 30 x 24
= 360 m
Total distance travelled → 300 + 360
= 660 m
Average Speed = 660 m / 40 s
= 16.5 m/s
c) See attached picture
d) Lorry travelled at constant speed, so for the entire 40 sec duration, it would have travelled:
25 m/s x 40 s = 1,000 m
We know that the car travelled a total of 660 m during the 40 sec, so the lorry would be (1,000 – 660) m ahead.
1000 – 660 = 340 m