Question

car is travelling with the speed of 36km/h the driver applies the brake retards the car uniformly the car is stopped in 5 second find:-
1. the retardation of the car 2.distance travelled before it stopped after applying the break

Answer 1

U = 36 Km/h = 10m/s.
V = 0
T = 5 seconds.
v= u + at
0 = 10 + a×5
-10 = 5a
-10/5 = a
-2 = a
so, retardation = 2m/s^2
By using 2nd equation of motion:-
S= ut + 1/2 at^2
s = 10×5 + 1/2 ×(-2)× 5×5
s = 50 + (- 25)
s = 25
Hope this helps frnd.

Answer 2

@
u= 36km/h= (36*5/18)m/s= 10m/s
v= 0
t= 5sec
_____
1) v=u+at
→0=10+a*5
→5a=-10
→a=(-10/5)
→a=-2m/s²
acceleration = -2m/s² so retardation = 2m/s²
_____
2) s=ut+1/2at²
→s=10*5+1/2*(-2)*(5)²
→s=50+1/2*(-50)
→s=50+(-25)
→s=25m
Distance travelled = 25metre
@:-)