Question

A man runs at a speed of $4 ms^{-1}$ to overtake a standing bus. When he is $6 m$ behind the door (at $t=0$ ), the bus moves forward and continues with constant acceleration of $1.2 ms^{-2}$ . The man shall access the door at time t is equal to:

Answer 1

Hint: In the question, initial velocity, initial distance, acceleration and final velocity are given. First, we found the distance for the man. Using and equating the second equation of motion, we get the quadratic equation in the variable of time. We will solve the equation by quadratic formula. We get the value of time.

Complete step-by-step solution: -
Initial velocity, $u = 0$
Initial distance, $d = 6m$
Final velocity, $v = 4 ms^{-1}$
Acceleration, $a = 1.2 ms^{-2}$
As we know,
Distance is equal to the product of velocity and time.
$s = vt$
Put values v and t
We get the distance for man.
$s = 4t$
Using the second equation of motion for man.
$s – d =ut + \dfrac{1}{2} a t^{2}$
Where, s and d are the final and initial distance.
$s =6 + \dfrac{1}{2} \times 1.2 \times t^{2}$
Put $ s = 4t$ in above equation
$4t = 6+ 0.6 t^{2}$
$ \implies 0.6 t^{2} – 4t +6 = 0$
We will solve this quadratic equation for getting the values of time. We will solve it by using a quadratic formula.
$t = \dfrac{-(-4) \pm \sqrt{\left( -4 \right)^{2} -4 \times 0.6 \times 6}}{2 \times 0.6}$
$\implies t = 2.3, 4.3$
At $t = 2.3 s$, the bus velocity is greater than man velocity so, man cannot reach the bus. But at $t = 4.3 s$, the relative velocity of the man and bus is zero so, man can catch the bus.
So, $t = 4.3 s$ is the right answer.

Note: The relative velocity is velocity of one body concerning another body. It is estimated as the rate of change of position of a body concerning another body. When both the objects are traveling in the same direction along parallel vertical lines with the same velocities, the velocity of each one concerning the other is zero.