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# Question

# Answer

a) Retardation of 1.4 m/s^{2} 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/s^{2} 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**

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