A car is moving with a velocity of 10 m/s.
Suddenly driver sees a child at a distance of 30 m
and he applies the brakes, it produced a uniform
retardation in car and car stops in 5 second.
Find (mass of car=1000 kg)
(1) the stopping distance of car
(ii) will the car hit the child
(iii) the retardation in car
(iv) resistive force on car
Answers
Answer:
2000 N.
Explanation:
S = 25 m
No
a = -2m/s²
F = -2000 N
Given :-
u = 10 m/s
Distance of child = 30 m
v = 0m/s
t = 5 s
To find :-
(i) the stopping distance of car
(ii) will the car hit the child
(iii) the retardation in car
(iv) resistive force on car
Solution:-
Since,the driver applied brakes and car retarted uniformly.
So , let's find the uniform retardation of car :-
→
→
→
Hence, the uniform retardation of car will be -2 m/s².
Now, find the distance travelled by car after applying the brakes.
→
→
→
→
Since,the driver applied brakes at 25 m but the child is at 30 m .
the driver applied brakes at 25 m but the child is at 30 m .hence, it stops before 5 m and the child is saved from accident.
We have to find the resistive frictional force in the car tyres.
m = 1000 kg
a = -2 m/s²
From Newton 2nd law of motion:-
→
→
The negative sign shows that force applied is opposite in direction of car.
hence, the magnitude of resistive force will be -2000 N.