2. A car is traveling with a speed of 22 m/s. The brakes can produce a maximum deceleration of 8.0 m/s. What is the minimum stopping distance for the car?
Answers
Answer:
30.25m is the required answer
Explanation:
According to the Question
it is given that
- Initial velocity ,u = 22m/s
- Final velocity ,v = 0m/s
- Deceleration ,a = 8m/s²
we need to calculate the minimum stopping distance for the car . So we apply here the kinematics equation .
- v² = u² + 2as
where,
v is the final velocity
a is the acceleration
u is the initial velocity
s is the minimum stopping distance
substitute the value we get
➾ 0² = 22² + 2×(-8) × s { we take acceleration negative.}
➾ 0 = 484 -16s
➾ -484 = -16s
➾ 484 = 16s
➾ s = 484/16
➾ s = 30.25m
- Hence, the minimum stopping distance for the car is 30.25metres.
Given :-
A car is traveling with a speed of 22 m/s. The brakes can produce a maximum deceleration of 8.0 m/s.
To Find :-
What is the minimum stopping distance for the car?
Solution :-
We know that
v² - u² = 2as
(0)² - (22)² = 2(-8)(s)
0 - 484 = -16s
-484 = -16s
-484/-16 = s
484/16 = s
30.25 = s
∴ 30.25 m is the minimum distance for stopping