Question

A car is travelling with a speed of 36km/h. The driver applies brakes retards the car uniformly. The car is stopped in 5seconds. Find the retardation of the car and the distance travelled before it is stopped.

Answer 1

Step 1: Given data
Initial speed, u=36 km/h = $36\times \frac{5}{18}$ = 10 m/s
Time, t=5 s
Final velocity, v=0 m/s

Let a and s be acceleration and total distance before stopping respectively.

Step 2: Formula used

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

$v^2=u^2+2as$ (Third equation of motion)

Step 3: Finding the retardation

From the equation of motion

$Or,0=10+a\times 5$

After solving

a = -2 m/$s^2$

Hence, the retardation is 2 m/$s^2$.

Step 4: Finding the distance traveled

From the equation of motion

$v^2=u^2+2as$

$Or,0={10}^2-2\times 2\times s$

After solving

s = 25 m

Hence, the distance traveled is 25 m.