Question

at the end of a race,a runner decelerates from a velocity of 9m/s at a rate of 2m/s^-2.
(a)how far does he travel in the next 5s?
(b)what is his final velocity ?
(c)how much time will it take to finally stop ?

Answer 1

Hii...
Here is your answer..

Solution:-
Initial velocity=9m/s
Acceleration= -2m/s^2
Distance travelled in 5 sec=?
We'll find it by using equation:-
$= > s = ut + \frac{1}{2} a {t}^{2} \\ = > s = 9 \times 5 + \frac{1}{2}( - 2) \times 5 \times 5$
$= > s = 45 + ( - 5) \times 5 \\ = > s = 45 - 25 \\ = > s = 20$
a) Distance travelled in 5 sec =20m
Now,
B) Final velocity=?
$= > {v}^{2} = {u}^{2} + 2as \\ = > {v}^{2} = {(9)}^{2} + 2 \times ( - 2) \times 20 \\ = > {v}^{2} = 81 - 80 \\ = > v = 1$
C)Time taken to stop=
$= > v = u + at \\ = > 1 = 0 + ( - 2)t \\ = > t = \frac{1}{2} = 0.5sec$
Hope it helps uh...✌️✌️✌️