Question

Speed of a car increases from 36 km/h to 90 km/h in 15 second find initial velocity and final velocity, acceleration and distance covered in 10 second

Answer 1

36km/h=10m/s=initial velocity (u)
90km/h=25m/s=final velocity (v)
time (t)=15 sec
a=(v-u)/t
=(25-10)/15
=15/15
=1m/s^2
:distance in 10 sec=36×10+1/2×3/5×10×10
=360+30
=390
:hope it's help.........mark as brainliest

Answer 2

Hii...
Here is your answer..

Solution:-
Initial velocity=36km/h
$= > 36 \times \frac{5}{18} \\ = > 10$
=>10m/s

Final velocity=90km/h
$= > 90 \times \frac{5}{18} \\ = > 25$
=>25m/s

Time =15 seconds

So,
$acceleration = \frac{v - u}{t} \\ = > \frac{25 - 10}{15} \\ = > \frac{15}{15} \\ = > 1$
Acceleration=1m/s^2

Now,
Distance travelled in 10 seconds,we will find it by using equation:-
$= > s = ut + \frac{1}{2} a {t}^{2} \\ = > s = 10 \times 10 + \frac{1}{2} \times 1 \times {(10)}^{2}$
$= > s = 100 + \frac{1}{2} \times 1 \times 100 \\ = > s = 100 + 50 \\ = > s = 150$
Hence,the Distance covered in 10 seconds=150m

Hope it helps uh...✌️✌️✌️