Question

a car travelling with a velocity of 80 km/h slowed down to 44 km/h in 15 s. The retardation is?

Answer 1

formula used (change in acceleration)
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Answer 2

Answer:

Acar traveling with a velocity of 80 km/h slowed down to 44 km/h in 15 s. The retardation is  $0.66666\ m/s^2$

Explanation:

Acceleration is given by the expression,

$a = \frac{v-u}{t}$                 ...(1)

where u - the initial velocity of the object,

           v - the final velocity and t is the time taken.

Or

Acceleration (a) of an object can be defined as the ratio of change in velocity and time.

In the question, it is given that,

$u = 80\ km/h = 80\times \frac{5}{18} = 22.2222\ m/s$

$v = 44\ km/h = 44\times \frac{5}{18} = 12.2222\ m/s$

$t = 15\ s$

Substitute these values into the equation(1)

$a = \frac{12.2222- 22.2222}{15} = -0.66666\ m/s^2$

The retardation of the car = $0.66666\ m/s^2$