Question

The retardation of a car that
comes to stop from a
velocity of 10 m/s in distance
of 50 m is​

Answer 1

Given :-

☄ Initial velocity of a car, u = 10 m/s

☄ Final velocity of the car, v = 0 m/s

☄ Distance travelled by the car, s = 50 m

To Find :-

retardation of the car

Solution :-

As we are provided with initial Velocity (u) , distance covered (s) and the car comes to rest .i.e., final velocity (v) is zero we can use 3rd equation of motion

So, using 3rd equation of motion .i.e.,

➠ v² - u² = 2as

here ,

v denotes final velocity, u denotes initial velocity, a denotes acceleration and s denotes distance.

by substituting all the given values,

➠ (0)² - (10)² = (2)(a)(50)

➠ 0 - 100 = 100a

➠ 100a = - 100

➠ a = -100/100

➠ a = - 1 m/s²

thus, retardation of the car is -1 m/s²

|| Note : if the speed is decreasing with time then accerlation is negative the negative accerlation is called retardation or deceleration ||

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