Question

plz solve 61 question i am unable to solve plz write whole steps

Answer 1

Suppose the bus is at origin.
Then the man is at 48 metres behind.

♦ Bus:

Initial velocity = 0 m/s
Acceleration a = 1 m/s²

So, its displacement can be written as a function of time as follows:

s = ut + ½ at²
So, s = 0 + ½ (1)t²
So, s = t²/2

Let's call this distance as Sb.
So, Sb = t²/2 metres

♦ Man:

Velocity = 10 m/s

So, the displacement which the man has can be written as a function of time as follows:

v = s/t
So, 10 = s/t
So, s = 10 t metres

But, the man is 48 metres behind the bus. So the displacement of man (Sm) can be written as:

Sm = 10 t - 48 metres

______________________

Now, when the man catches the bus, both will have the same coordinates. In other words, both will have same displacement.

So, Sb = Sm
So, t²/2 = 10t - 48
So, t² = 20t - 96
So, t² - 20t + 96 = 0
So, t² -12t - 8t + 96 = 0
So, t(t-12) -8(t-12) = 0
So, (t-12)(t-8) = 0

So, t-12 = 0 or t-8=0
So, t = 12 s or t = 8 s

(It means that if the man keeps running at 10 m/s, then he would overtake the bus at t = 8s. But the bus is accelerating. So, eventually the bus would overtake again at t = 12 s)


Here, the man wants to catch the bus, which he can do at a minimum time of t = 8s


So, your answer is 8 seconds