Question

A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/s2, at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus?
​

Answer 1

SOLUTION :

GIVEN :

Bus is at rest . So , here

u = 0 m/s

a = 1 m/s²

Bus covers distance 's' in 'n' seconds. Now , show this in the form of

$\mathsf{s = ut + \frac{1}{2} a {t}^{2}}$

Substitute the values in the above formula

$\mathsf{s = 0t + \frac{1}{2} {n}^{2}}$

$\mathsf{So ,\: s_{bus} = \frac{ {n}^{2} }{2}}$

Here , man is running with uniform velocity of 10 m/s.

v = 0 m/s

t= 'n' seconds

Distance = speed × time

So , s= 10n metres.

After 'n' seconds , man catches the bus.

So ,

$\mathsf{s_{man} = 48 + s_{bus}}$

$\mathsf{10n = 48 + \frac{ {n}^{2} }{2}}$

$\mathsf{20n = 96 + {n}^{2} }$

$\mathsf{{n}^{2} + 96 - 20n = 0}$

$\mathsf{(n - 12)(n - 8) = 0}$

They asked to find the minimum time , so minimum time is 8 seconds.

NOTE :

If the body is in rest, take initial velocity as zero.

Use correct formulas.

Multiply the variables and numbers correctly.

Some formulas for uniform accelerated motion are ,

$\mathsf{s = ut + \frac{1}{2} a {t}^{2}}$

$\mathsf{v = u + at}$

$\mathsf{{v}^{2} - {u}^{2} = 2as}$

Answer 2

read it.....

hope it is helpful......

please if it is helpful then mark it as brainlist as I really need it plz a humble request to you..........

and please follow me............