Question

a bus starts to move with an acceleration of 1ms-2.a man who is 48m behind the bus runs with constant velocity od 10ms-1to catch it.in how much time he will catch the bus​
with prcedure

Answer 1

Explanation:

acceleration (a) = 1ms-2

distance = 48 m

velocity(v) = 10ms-1

time=?

1. v=at

10=1*t

10 sec =t

Hope itis right

Answer 2

Answer:

ANSWER is 6s

Explanation:

Velocity of man = 10 m/s

S = 42 m

Acceleration of man with respect to bus = Acceleration of man - Acceleration of bus = 0 - (1) = -1 m/s

Applying second equation of motion,

s = Vt + 1/2 at2

42= 10t - 1/2*t2

42= (20t-1t^2)/2

84= 20t - 1t^2  

1t^2 - 20t +84 = 0

It is in the form of quadratic equation..  

t^2-6t-14t+84=0

t(t-6)-14(t-6)=0

(t-6)(t-14)=0

t-6=0;t-14=0

Therefore t=14s and t=6s

We need minimum value

So we take 6s.