Question

A car is travelling to the right with a speed of 42\,\dfrac{\text m}{\text s}42 s m ​ 42, space, start fraction, m, divided by, s, end fraction when the driver slams on the brakes. The car skids for 4.0\,\text s4.0s4, point, 0, space, s with constant acceleration before it comes to a stop. How many meters did the car skid before coming to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

Answer 1

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed $v_i = 42 m/s$

final speed $v_f = 0$

time taken by the car to stop

$t = 4s$

now the distance moved by the car before it stop is given as

$d = \frac{v_f + v_i}{2} \times t$

now we have

$d = \frac{42 + 0}{2} \times 4$

$d = 84 m$

Answer 2

Answer:

84

Explanation:

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