Question

a motorbike initially moving at theach 18km/h accelerates at the rate of 5m/s square for 5 seconds. calculate the distance covered by it from start and final velocity

Answer 1

Hiii...
Here is your answer..

Solution:-
Initial velocity= 18km/h
$= > 18 \times \frac{5}{18} \\ = > 5m \: per \: sec$
=>5m/s

Acceleration= 5m/s^2
Time =5 seconds
Distance travelled=?

By using equation,that is,
$= > s = ut + \frac{1}{2} a {t}^{2}$
$= > s = 5 \times 5 + \frac{1}{2} \times 5 \times 5 \times 5 \\ = > s = 25 + \frac{1}{2} \times 125 \\ = > s = 25 + \frac{125}{2}$
$= > s = \frac{50 + 125}{2} \\ = > s = \frac{175}{2} \\ = > s = 87.5$
Therefore,
=>Distance travelled=87.5m

Now,
We have to find final velocity,we'll find it by using equation:-
v=u+at
v=5+87.5×5
v=442.5

Hence,the final velocity=442.5m/s

Hope it helps uh.....✌️✌️✌️