Question

The brakes applied to a car produce an acceleration of 6 ms-² in opposite direction to the motion. If the car takes 2 s to stop after the application of rakes, calculate the distance it travels during this time.

Answer 1

Step 1: Given data

Acceleration(a) = -6 m/$s^2$ (as brakes applied)

Time taken(t) = 2 sec

Final velocity(v) =0

Let 's' is the distance traveled and 'a' is acceleration.

Step 2: Formula used

$v=u+at$ (First equation of motion)

$s=ut+\frac{1}{2}a{t}^2$ (Second equation of motion)

Step 3: Finding initial velocity

$v=u+at$

Putting all the values

$0=u+(-6)\times 2$

Or, u = 12 m/s

Hence the initial velocity is 12 m/s.

Step 4: Finding the distance traveled

$s=ut+\frac{1}{2}a{t}^2$

Putting all the values

$s=12\times 2-\frac{1}{2}\times 6\times 2\times 2$

After solving

s = 12 m

Hence, the distance traveled = 12 m.