Question

A man runs at a constant speed of 4 m/s to overtake a standing bus. When
he is 6.0 m behind the door (t=0), the bus moves forward and continues
with a constant acceleration of 1.2 m/s2.
a) How long does it take for the man to reach the door?
b) If in the beginning he is at 10.0 m from the door of the buss, will he
ever catch up?

Answer 1

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Answer 2

Answer: (a).The man will reach in 2.3 s and (b). he will not catch the bus

Explanation:

Given that,

Speed of man v= 4 m/s

Distance d = 6.0 m

Acceleration $a=1.2m/s^2$

(a). Using formula of distance

For man,

$s = v\times t$

$s = 4t$....(I)

Using equation of motion

For bus,

$s = d+ut+\dfrac{1}{2}at^2$

$s = 6+\dfrac{1}{2}\times1.2t^2$.....(II)

Equate the equation (I) and (II)

$4t-6=\dfrac{1}{2}\times1.2t^2$

$0.6t^2-4t+6=0$

$t_{1} =4.4\ s and t_{2}= 2.3\ s$

The man will reach in 2.3 s

(b). If i the beginning he is at 10.0 m from the door of the bus

Using equation of motion

$s = d+ut+\dfrac{1}{2}gt^2$

$s = 10+\dfrac{1}{2}at^2$.....(II)

Equate the equation (I) and (II)

$4t-10=\dfrac{1}{2}\ties1.2t^2$

$0.6t^2-4t+10=0$

$t_{1} = \dfrac{10}{3}+i\dfrac{5\sqrt{2}}{3}\ s and t_{2}=\dfrac{10}{3}-i\dfrac{5\sqrt{2}}{3}\ s$

The time can not determine for this case.

Therefore, he will not catch the bus.

Hence, This is the required solution.