The brake applied to a car produce an acceleration of 6m/s^2 in opposite direction. If the car stops in 2 seconds calculate the distance travelled

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Solution

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

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The brake applied to a car produce an acceleration of 6m/s^2 in opposite direction. If the car stops in 2 seconds calculate the distance travelled

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

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