Question

A car moving on a straight road covers one fourth of the total distance with a constant speed of 90km/hr with what speed it must cover the remaining distance such that the average speed becomes 120km/hr for entire journey?

Answer 1

Let the total distance be $d$ and let the speed with which it covers the remaining distance be $v$. Now,$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\\\Rightarrow 120=\frac{d}{t_1+t_2}\\\Rightarrow 120=\frac{d}{\frac{\frac{d}{4}}{90}+\frac{\frac{3d}{4}}{v}}}\\\Rightarrow \frac{1}{120}=\frac{1}{360}+\frac{3}{4v}\\\Rightarrow \frac{1}{180}=\frac{3}{4v}\\\Rightarrow v=135 \frac{km}{h}$

Answer 2

Answer:The speed is 135\ km/h it must cover the remaining distance.

Explanation:

Given that,

Speed = 90 km/h

Average speed = 120 km/h

let us considered the total distance is d.

We know that,

Average velocity = Total distance /total time

$v_{avg} =\dfrac{d}{T}$

Where, T = total time = $t_{1}+t_{2}$

$120\ km/h= \dfrac{d}{t_{1}+t_{2}}$

$120\ km/h = \dfrac{d}{\dfrac{d}{4\times 90}+\dfrac{3d}{4\times v}}$

Where, v is the velocity it must cover the remaining distance.

$120\ km/h= \dfrac{1}{\dfrac{1}{360\ km/h}+\dfrac{3}{4v}}$

$v = 135\ km/h$

Hence, The speed is 135\ km/h it must cover the remaining distance.