Motion — Question 6

(a) In first part of journey
speed = 30 km h^-1

time = 30 min
60 min = 1 h
30 min = $\dfrac{\text{1}}{\text{2}}$ h

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] \text{30} = \dfrac{\text{distance}}{\dfrac{1}{2}} \\[1em] \text{distance} = 30 \times {\dfrac{1}{2}} = 15 \text{km}$

In second part of journey
speed = 40 km h^-1

time = 1 h

$\text{Speed} = \dfrac{\text{distance}}{\text{time}} \\[1em] \text{40} = \dfrac{\text{distance}}{\text{1}} \\[1em] \text{distance} = 40 \times 1 = 40 \text{km}$

So, total distance travelled = 15 km + 40 km = 55 km.

(b) time taken in first part of journey = 30 min

60 min = 1 h
30 min = $\dfrac{\text{1}}{\text{2}}$ h

time taken in second part of journey = 1 h

total time taken = 1 h + $\dfrac{\text{1}}{\text{2}}$ h = 1.5 h

So, total time taken = 1.5 h.

(c) $\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \\[1em] \text{Average Speed} = \dfrac{\text{55}}{\text{1.5}} = 36.67 \text{km h}^{-1}$

So, average speed of the car = 36.67 km h^-1.