Question

A Car is moving at a speed of 72 Km/h. Brakes are applied to stop it in 4 seconds. Find
a. the retardation produced and
b. the distance travelled by the car before coming to rest.
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Answer 1

Given:

Initial velocity, u = 72 km/h

Final velocity, v = 0 km/h

Time taken, t = 4s

To find:

i) the retardation produced

ii) the distance travelled by the car before coming to rest.

Answer:

i)

$a = \frac{v - u}{t}$

$a = \frac{0 - 72}{4}$

$a = - 18m \: {s}^{ - 2}$

So, the retardation is 18 m/s^2 (Retardation is negative acceleration)

ii) By using the formula s = ut+ 1/2at^2

$s = 72 \times 4 + \frac{1}{2} \times - 18 \times 4$

s = 288 + (-36)

s= 252 m

So, the distance covered is 252 m

Answer 2

Answer:

t square is 16 but you have taken 4 only