Question

The speed of a car is reduced from 90 km hr to 36 km hr in 5 s. What is a distance travelled by the car during this time interval

Answer 1

initial velocity (u) = 90km/h.
= 25m/s
final velocity (v) = 36 km/h
= 10m/s

time taken = 5 s.

acceleration (a)= (v-u)/t
= ( 10 - 25)/ (5)
= -15/5
= -3m/s^2


let ,
distance travelles be s

Using third equation of motion

2as =v^2 - u^2
2 (-3)(s) = (10)^2 -(25)^2
-6s = 100 - 625
- 6s = - 525
6s = 525
s = 525/6
= 87.5m

Answer 2

Heya friend,

Note-

u=initial velocity
v=final velocity
t=time
s=Distance travelled
a=Acceleration

u= 90km/hr = 25m/s
v= 36km/hr= 10m/s
t=5s
s=?

a=?
a=v-u/t
a= 10-25/5
a=-3m/s^2

We can use the third equation of motion-

2as= v^2-u^2
2×-3×s= (10)^2-(25)^2
-6s= 100-625
-6s=-525
6s=525
s= 525/6
s=87.5m


Hope it helps...!!