Question

Can someone help me out with this?

Answer 1

Answer:

Here, Total distance = 100 m

For the first half of run, Distance = 50 m

u = 0 m/s , a = 1 m/s^2

so ,

s = ut+1/2 x a x t^2

50 = 0+1/2 x 1 x t^2

t = 10 sec

Now,

At the end of 50 m,

v = u+at = 0+ 1 x 10 = 10 m/s

For the second half of run, we calculate for s=25 m

first(uniformly accelerated motion)

Here, u = 10 m/s , a = 1 m/s^2

So, v^2 = u^2 + 2 x a x s = 100 + 2 x 1 x 25 = 150

v = 12.24 m/s

now,

for time t,v = u + at => 12.24 = 10 + t

t = 2.24 sec

For the last 25 m :-

t = distance/speed = 25/12.24=2.04 sec

Hence, Time taken for the first half of run(50 m) = 10 sec    and,

Time taken for the second half of run(50 m) = 2.24 + 2.04 = 4.28 sec