Brakes applied to car produce an acceleration of 6m/s^2 in opposite direction to motion if the car takes 2s to stop after application of brakes then calculate the distance it travells during this time
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Answered by
5
Here ,
u(initial velocity) = ?
v(final velocity) = 0m/s
t (Time) = 2 seconds
a (Acceleration) = -6m/s²
So , Use the formula of kinematics to find u.
v = u + at
or, 0 = u + (-6)(2)
or, 0 = u + (-12)
or, 0 = u - 12
or, 12 = u
or, u = 12
Now , we have got Initial velocity , To find distance (s) , we will again apply formula of Kinematics→
s = ut + (1/2)at²
or, s = (12)(2) + (1/2)(-6)(2)²
or, s = 24 + (1/2)(-24)
or, s = 24+(-12)
or, s = 12m
Answer → The distance travelled by car is 12m.
____________________________
u(initial velocity) = ?
v(final velocity) = 0m/s
t (Time) = 2 seconds
a (Acceleration) = -6m/s²
So , Use the formula of kinematics to find u.
v = u + at
or, 0 = u + (-6)(2)
or, 0 = u + (-12)
or, 0 = u - 12
or, 12 = u
or, u = 12
Now , we have got Initial velocity , To find distance (s) , we will again apply formula of Kinematics→
s = ut + (1/2)at²
or, s = (12)(2) + (1/2)(-6)(2)²
or, s = 24 + (1/2)(-24)
or, s = 24+(-12)
or, s = 12m
Answer → The distance travelled by car is 12m.
____________________________
Answered by
6
12 m
Explanation:
Acceleration produced on applying brakes
= -6 m/(s)^2
Time taken to stop = 2 s
Let initial velocity be u m/s
final velocity is 0 m/s
Using equation of motion
v = u + at
0 = u +(-6)×(2)
u = 12 m/s
Now, using
(v)^2 - (u)^2 = 2as
(0)^2 - (12)^2 = 2 (-6)(s)
-144 = -12(s)
s = 12 m
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