Question

a car accelerates uniformly from 27 km per hour to 45 km per hour in 5 seconds calculate the acceleration and the distance covered by car in that time

Answer 1

Initial velocity (u) =  27 km/h = 7.5 m/s

Final velocity (v) = 45 km/h = 12.5 m/s

Time taken = 5 seconds

Let the acceleration be a.

Now,

$v = u + at$

$a = \frac{12.5 - 7.5}{5}$

a = 1 m/s^2

Hence, the acceleration is 1 m/s^2

Now, Let the distance covered be s.

$v^2 = u^2 + 2aS$

$s = \frac{12.5^2-7.5^2}{2}$

s = 50 m

Hence, the distance covered is 50 m.

Answer 2

u is 27 km per hr 27*1000/3600 7.5m/s
v is 45 km per hr 45*1000/3600 12.5m/s
t is 5 sec
a = v-u/t
a= 12.5-7.5/5
a=5/5m/s²
a=1 m/s²
s=ut +½at²
7.5*5+½*1*5*5
37.5+25/2
37.5+12.5
50m
hope this helps you