The brakes applied to a car produces an aceleration of 6m/s square in the opposite direction to the motoin. If the car takes two seconds to stop after the application of brakes then calculate the distance it travels during this time
Answers
Answered by
1
Answer:
Acceleration a = -6 m/s^2
Time t = 2 s
Final velocity v = 0 m/s
Let initial velocity be u
Let distance be s
v = u + at
So, 0 = u + (-6)(2)
So, u = 12 m/s
Now, s = ut + (1/2) at^2
So, s = 12(2) + (1/2)(-6)(2^2)
So, s = 24 - 12
So, s = 12 m
Thus, distance travelled is 12 m
Answered by
8
Answer:
Hey mate....
Here's ur answer :::::::::::::""""
Given : Acceleration (a) = –6m/s²
time (t) = 2s
final velocity (v) = 0m/s
We know that ,
v = u + at
0 = u + (–6) × 2
.°. u = 12 m/s
Also,
s = ut + 1/2 at²
.•. s = 12 * 2 + 0.5 * (–6) * 2²
, s = 24 – 12
.°. s = 12 m.
Thus, the car will move 12 m before it stops after the application of brakes.
Hope it helps!!!
❤❤❤❤❤
Plz! Mark me as brainliest.....
and don't forget to follow me... ✌✌
Similar questions