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Physics

A car travels first 30 km with a uniform speed of 60 km h-1 and then next 30 km with a uniform speed of 40 km h -1.

Calculate —

(i) the total time of journey,

(ii) the average speed of the car.

Motion in One Dimension

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Answer

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,

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

Substituting the values in the formula above, we get,

t1=S1v1t1=30km60km h1t1=12ht1=0.5h\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,

t1=0.5×60 mint1=30 min\Rightarrow \text{t}1 = 0.5 \times 60 \text{ min} \\[0.5em] \Rightarrow \text{t}1 = 30 \text{ min} \\[0.5em]

and

t2=S2v2t2=30km40km h1t2=34ht2=0.75h\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,

t2=0.75×60 mint2=45 min\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,

Average speed=total distancetotal time takenAverage speed=60 km1.25 hAverage speed=48 kmh1\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

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