The brakes applied to a moving car produce a negative acceleration of 6m/s^2. If the car takes 2s to stop after applying brakes, calculate the distance it travels during this time.the anser is 12m but i need the steps
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146
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
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
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