The brakes applied to a car produce an acceleration of 6 ms-2 n the opposite direction of the motion. If the car take 2s to stop after the application of brake, calculate the distance traveled by the car during this time.
Answers
Answered by
2
v=0 because after applying brakes the car comes to rest.
t=2 s
a=-6ms⁻²
u=?
So, we have to find the initial velocity of the car.
For this we have to use the first equation of motion. We have;
v=u+at
0=u+(-6)x2
0=u-12
u=12 m/s
Now to find the distance we can use either the second or third equation of motion.
I will show you in both ways
By using the second equation, we have:
s=ut+1/2.at^2
24+1/2.(-6)x4
24-12=12m
By using the third equation, we have:
v²-u²=2as
0-144=2x(-6)xs
-144=-12s
12s=144
s=12m
Original Credit to Birendrak1975
t=2 s
a=-6ms⁻²
u=?
So, we have to find the initial velocity of the car.
For this we have to use the first equation of motion. We have;
v=u+at
0=u+(-6)x2
0=u-12
u=12 m/s
Now to find the distance we can use either the second or third equation of motion.
I will show you in both ways
By using the second equation, we have:
s=ut+1/2.at^2
24+1/2.(-6)x4
24-12=12m
By using the third equation, we have:
v²-u²=2as
0-144=2x(-6)xs
-144=-12s
12s=144
s=12m
Original Credit to Birendrak1975
Answered by
0
v = u + at
0 = u + (-6 × 2)
u = 12 m/s
S = u^2 / (2a)
= 12^2 / (2 × 6)
= 12 m
Distance travelled by car during this time is 12 m
0 = u + (-6 × 2)
u = 12 m/s
S = u^2 / (2a)
= 12^2 / (2 × 6)
= 12 m
Distance travelled by car during this time is 12 m
Similar questions