Motion in One Dimension — Question 6

Given,

Distance travelled S₁ = 30 km

Speed v₁ = 60 km h^-1

Distance travelled S₂ = 30 km

Speed v₂ = 40 km h^-1

(i) As we know,

$\text {Time (t)} = \dfrac {\text {distance (S)}}{\text {speed (v)}}$

Substituting the values in the formula above, we get,

$\text{t}_1 = \dfrac{\text{S}_1}{\text{v}_1} \\[0.5em] \text{t}_1 = \dfrac{30 \text{km}}{60 \text{km h}^{-1}} \\[0.5em] \Rightarrow \text{t}_1 = \dfrac{1}{2} \text{h} \\[0.5em] \Rightarrow \text{t}_1 = 0.5 \text{h} \\[0.5em]$

Converting 0.5 h to min we get,

$\Rightarrow \text{t}_1 = 0.5 \times 60 \text{ min} \\[0.5em] \Rightarrow \text{t}_1 = 30 \text{ min} \\[0.5em]$

and

$\text{t}_2 = \dfrac{\text{S}_2}{\text{v}_2} \\[0.5em] \text{t}_2 = \dfrac{30 \text{km}}{40 \text{km h}^{-1}} \\[0.5em] \Rightarrow \text{t}_2 = \dfrac{3}{4} \text{h} \\[0.5em] \Rightarrow \text{t}_2 = 0.75 \text{h} \\[0.5em]$

Converting 0.75 h to min we get,

$\Rightarrow \text{t}_2 = 0.75 \times 60 \text{ min} \\[0.5em] \Rightarrow \text{t}_2 = 45 \text{ min} \\[0.5em]$

Total time t = t₁ + t₂
= 30 min + 45 min
= 75 min

Hence, Total time of journey = 75 min.

(ii) As we know,

Total distance S = S₁ + S₂
= 30 km + 30 km
= 60 km

Hence, Total distance travelled = 60 km.

As we know,

$\text{Average speed} = \dfrac{\text{total distance}}{\text{total time taken}} \\[0.5em] \text{Average speed} = \dfrac{60 \text { km}}{1.25 \text { h}} \\[0.5em] \Rightarrow \text{Average speed} = 48 \text { kmh} ^{-1}$

Hence, Average speed of the car = 48 kmh^-1