Question

A bus starts to move with an acceleration of 1 m/s ² . A man who is 48m behind the bus runs with constant velocity of 10 m/s to catch it. In how much time he will catch the bus?​

Answer 1

Question:-

• A bus starts to move with an acceleration of 1 m/s² . A man who is 48m behind the bus runs with constant velocity of 10 m/s to catch it. In how much time he will catch the bus ?

Given:-

• A bus starting from rest and moving with and acceleration of 1 m/s² and 48 m behind a man round with constant velocity of 10 m/s to catch the bus .

To Find:-

• Find the time taken by the man to catch the bus .

Solution:-

Here ,

$\tt\implies \: v = t$

$\tt\implies \: 10t = 48 + \dfrac { { t }^{ 2 } } { 2 }$

$\tt\implies \: 20t = 96 + { t }^{ 2 }$

$\tt\implies \: { t }^{ 2 } - 20t + 96 = 0$

$\tt\implies \: { t }^{ 2} - 12t - 8t + 96 = 0$

$\tt\implies \: t ( t - 12 ) - 8 ( t - 12 ) = 0$

$\tt\implies \: ( t - 12 ) ( t - 8 ) = 0$

$\tt\implies \: t = 8 , 12$

Here ,

• The man catches the bus at 8 sec .