Question

a car travelling at 22.4m/s skids to stop 2.55s determine the skidding distance of the car(assume uniform acceleration)

Answer 1

A car is traveling at a speed of  22.4 m/s and skids to a stop in 2.55 s. So we have following information:

Initial speed of the car i.e  u= 22.4 m/s

Final speed of the car  i.e.  v=0 m/s  ( because car stops)

Time taken to come to stopping position  t= 2.55 sec

Answer 2

Given,

A car is traveling at 22.4m/s and skids to stop at 2.55s.

To Find,

The skidding distance =?

Solution,

We can solve the question as follows:

It is given that a car is traveling at 22.4m/s and skids to stop at 2.55s. We have to find the skidding distance.

$Initial\: velocity = 22.4\: m/s$

$Final\: velocity = 0\: m/s$      (Since it stops after 2.55 seconds)

$Time = 2.55\: s$

First, we will find the acceleration of the car.

According to the first equation of motion,

$v = u + at$

Where,

$v = Final\: velocity\\u = Initial\: velocity\\t = Time\\a = Acceleration$

Substituing the values in the above formula,

$0 = 22.4 + a*2.55$

$-22.4 = a*2.55$

$a = \frac{- 22.4}{2.55} = -8.78\: m/s^{2}$

The acceleration is negative because it is in the opposite direction of the velocity.

Now, we will find the skidding distance.

According to the third equation of motion,

$v^{2} - u^{2} = 2as$

Substituting the values in the above formula,

$0 - (22.4)^{2} = 2*(-8.78)*s$

$-(22.4)^{2} = -2*8.78*s$

$s = \frac{(22.4)^{2} }{2 * 8.78}$

$s = 28.57\: m$

Hence, the skidding distance is 28.57 m.