Question

A man is $48\;m$ behind a bus which is at rest. The bus starts accelerating at the rate of $1\;m{s^{ - 2}}$, at the same time the man starts running with uniform velocity of $10\;m{s^{ - 1}}$. What is the minimum time in which the man catches the bus?

Answer 1

Hint: The man is moving towards bus with constant velocity, and the bus is moving with constant acceleration, in both cases acceleration is either zero or constant so we can apply Newton’s equation of motion which can relate acceleration velocity and time with each other, to find out the time at which man catches the bus.

Formula used:
$S = \dfrac{1}{2} \times a \times {t^2}$
$S$ Is displacement
$a$ Is acceleration
$t$ Is time

Complete step-by-step answer:
Let the man catches the bus at time $t$, in this time the bus would cover a displacement of$S\;m$, since initially the man was $48\;m$ behind the bus the man would cover a displacement of $\left( {S + 48} \right)\;m$.
To cover a displacement of$S\;m$, it takes $t$ time by the bus so applying second equation of motion, we get,
$
  S = \dfrac{1}{2} \times a \times {t^2} \
  S = \dfrac{1}{2} \times 1 \times {t^2} \
 $
Now, for man to cover a displacement of $\left( {S + 48} \right)\;m$ it also takes $t$ time, than applying Newton’s equation of motion’
$
  \left( {S + 48} \right) = 10 \times t \\
  S = 48 + 10 \times t \\
 $
We see that the LHS of these two equations are the same so we can equate their RHS with each other.
$
  \dfrac{1}{2} \times 1 \times {t^2} = 48 + 10 \times t \
  \dfrac{{{t^2}}}{2} - 48 - 10 \times t = 0 \
 $
On multiplying the whole equation by $2$ we get,
${t^2} - 20 \times t - 96 = 0$, for solving this quadratic equation we can use quadratic formula to find $t$
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$a = 1,\;b = - 20,\;c = - 96$
$
  t = \dfrac{{ - \left( { - 20} \right) \pm \sqrt {{{\left( {20} \right)}^2} - 4 \times 1 \times \left( { - 96} \right)} }}{{2 \times 1}} \
  t = \dfrac{{20 \pm \sqrt {400 - 384} }}{2} \
 $
$
  t = \dfrac{{20 \pm 4}}{2} \
  t = 12\sec ,\;8\sec \
 $
Since we want minimum time so that man catches the bus we reject the value $12\;\sec $ and accept the value $8\;\sec $.
So the minimum time in which the man catches the bus is $8\;\sec $.

Note: Be careful of the fact that Newton’s equation of motion can only be used if acceleration of any body is constant or zero.
Aliter: The situation can also be solved using the concept of relative motion in which we can solve the complete question from the frame of reference of the bus or from frame of reference of man.