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# Question 36

A car travels the first 30km with a uniform speed of 60km h-1 and then next 30km with a uniform speed of 40 km h-1. Calculate: (i) the total time of the journey, (ii) the average speed of the car.

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Given: Let t1, d1 and s1 be the time, distance and speed travelled by car in the first part of the journey. Let t2, d2 and s2 be the time, distance and speed travelled by car in the second part of the journey.

\begin{array}{l} \mathrm{d} 1=30 \mathrm{km}, \mathrm{sl}=60 \mathrm{km} \mathrm{h}^{-1}, \mathrm{t} 1=? \\ \mathrm{d} 2=30 \mathrm{km}, \mathrm{s} 2=40 \mathrm{km} \mathrm{h}^{-1}, \mathrm{t} 2=? \end{array}

(i) Total time of the journey = t1 + t2 We know that, \text { Speed }=\frac{\text { distance }}{\text { time }} \Rightarrow \text { time }=\frac{\text { distance }}{\text { speed }}

\begin{aligned} \Rightarrow t 1 &=\frac{d 1}{s 1} \\ &=30 / 60=1 / 2 \mathrm{hr}=0.5 \mathrm{hr} \\ \Rightarrow \mathrm{t} 2 &=\frac{\mathrm{d} 2}{\mathrm{s} 2} \\ &=30 / 40=0.75 \mathrm{hr} \end{aligned}

Total time = t1 + t2 = 0.5 + 0.75 = 1.25hrs or 75 minutes

(ii) \text { Average speed }=\frac{\text { total distance travelled }}{\text { total time taken }} \qquad=\frac{d 1+d z}{t 1+t 2}=\frac{30+30}{1.25}=\frac{60}{1.25}=48 \mathrm{km} \mathrm{h}^{-1}

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