Question

A boy runs at a constant speed of 2 ms–1 to catch a

bus standing at the bus stop. When the boy was 4.2

m behind the bus, it suddenly started with an accel-

eration of 0.4 ms–2

.

(i) What further time will the boy take to catch

the bus?

(ii) If the boy does not catch the bus but continues

to run at the same rate, at what time will the

bus overtake the boy?

(iii) If the boy was 8 m behind the bus when it started

with an acceleration of 0.4 ms–2, could the boy

had been sucessful in catching the bus?​

Answer 1

Given:

Boy was running at 2 m/s. When the boy was 4.2 m behind the bus , the bus started with an acceleration of 0.4 m/s².

To find:

• Further time after which boy catches bus

• Time after bus overtakes boy , if boy kept running.

• If the boy was 8 m , could the boy catch the bus ?

Calculation:

Let the time taken for the boy to catch bus be t:

So in time t , the boy and the bus travels same distance ;

$\therefore \: 2t = 4.2 + \bigg \{ \dfrac{1}{2} \times (0.4) \times {t}^{2} \bigg \}$

$= > 2t = 4.2 + 0.2 {t}^{2}$

Multiplying in both sides with 10 ;

$= > 20t = 42 + 2 {t}^{2}$

$= > 2 {t}^{2} - 20t + 42 = 0$

$= > 2 {t}^{2} - 14t - 6t + 42 = 0$

$= > 2t(t - 7) - 6(t - 7) = 0$

$= > (2t - 6)(t - 7) = 0$

Either , 2t - 6 = 0 => t = 3 sec OR

t - 7 = 0 => t = 7 sec.

So the boy catches the bus for the first time at 3 seconds.

If he kept running , the bus would have overtaken him in 7 seconds.

If the boy had been 8 m away when bus started , the Equation would have been :

$\therefore \: 2t = 8 + \bigg \{ \dfrac{1}{2} \times (0.4) \times {t}^{2} \bigg \}$

$= > 2t = 8 + 0.2 {t}^{2}$

$= > 20t = 80 + 2 {t}^{2}$

$= > 2 {t}^{2} - 20t + 80 = 0$

Calculating the Discriminant of this quadratic equation , we get :

$= > D = {b}^{2} -4ac$

$= > D = {(20)}^{2} -( 4 \times 2 \times 80)$

$= > D = 400-640$

$= > D = - 240$

$= > D < 0$

Since Discriminant is negative , the quadratic equation has no real roots , hence the boy couldn't catch the bus.