A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate distance covered by the bus in metres during this interval.

Here initial velocity $u=54\ km/h=54\times\frac{5}{18}=15\ m/s$

Final velocity of the bus $v=72\ km/h=72\times\frac{5}{18}=20\ m/s$

Time $t=10\ sec$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{20-15}{10}$

$=\frac{5}{10}\ m/s^2$

$=0.5\ m/s^2$

Therefore, the acceleration is $0.5\ m/s^2$

Let $s$ be the distance travelled by the bus.

On using the equation of motion $v^2=u^2+2as$

$20^2=15^2+2\times0.5\times s$

Or $400=225+s$

Or $s=400-225$

Or $s=175\ m$

Therefore, the distance travelled by the bus is $175\ m$.

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Updated on: 10-Oct-2022

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