Question

A particle initially at rest , moves with an acceleration 5m/s square for 5 sec . Find distance travelled in 4 s ,5s and 5 th second

Answer 1

h= ut+1/2at^2
h1 = 0 + 1/2 (5) (4) ^2
h1=5*8
h1=40m

h2= 0 + 1/2 (5)(5)^2
h2= 25m

Answer 2

Answer:

The distance traveled in  $5^{th}\ s$ is 22.6 m

Explanation:

Given that,

Acceleration $a = 5 m/s^2$

Time t = 5 sec

We need to calculate the distance

Using equation of motion

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

Put the value in equation

$s=0+\dfrac{1}{2}\times5\times5^2$

$s=62.5\ m$

Now, the distance traveled in 4 second,

Using equation of motion again

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

$s=0+\dfrac{1}{2}\times5\times4^2$

$s = 40\ m$

The distance traveled in $5^{th}\ sec$

$s_{5}=s_{5}-s_{4}$

$s_{5}=62.6-40$

$s_{5}=22.6\ m$

Hence, The distance traveled in  $5^{th}\ s$ is 22.6 m