Question

on
two
3 A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate
of 1 m/s, at the same time the man starts running with uniform velocity of
10 m/s. What is the minimum time in which the man catches the bus ? (8)
AS, 4 Marks
A Bus is at rest.​

Answer 1

Answer:

Bus is at rest. <br>

<br> Let the bus covers the distance 's' in 'n' seconds <br>

<br>

<br> A man running with uniform velocity, v = 10m/sec. <br> Distance covered by man in n seconds = 10 nm. <br> But after 'n' seconds the man catches the bus. <br>

After n seconds,

<br>

<br>

The minimum time in which the man catches the bus is 18 sec.

Answer 2

Answer:

The bus was at rest then initial velocity u=0

acceleration a=0

therefore final velocity that is its speed will be

v=at

speed of bus bus will increase by 1m/s for every second

speed of man =10 m/s

therefore in 5 sec

the man will be 50 m from his original position and bus will be 5 meters

distance between them will be

53-50 = 3 m

therefore the man will catch the bus in

if 10 m= 1 sec

then 3m= 0.1 sec

total time taken by man to catch the bus = 5.1 secs