Question

Hello! Please solve this numerical.

An object is pushed along a horizontal surface in such a way that it starts with a velocity of 36km/h and decreases at the rate of 0.5m/s^2. The time it will take to come to rest, is...?

Answer 1

⭐⭐⭐⭐⭐ ANSWER ⭐⭐⭐⭐⭐

Given, initial velocity (u) = 36 km/h = 10 m/s.

retardation (r) = -0.5 m/s²

and its final velocity (v) will ultimately comes to zero amd hence, we can say, v = 0 m/s.

Now, r = (v-u)/t

=> -0.5 = (0-10)/t

=> -0.5t = -10

=> t = 10/0.5

=> t = 20 seconds

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✌✌✌✌ OUTCOMES ✌✌✌✌

1] WHY WE ARE HOLDING "v'' = 0 m/s?

Ans : Because since the velocity keeps decreasing at the rate of 0.5 m/s², ultimately, the body should come to rest and it means that its final velocity will be zero. Hence, "v" = 0 m/s.

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⭐⭐⭐ ALWAYS BE BRAINLY ⭐⭐⭐

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Answer 2

u = 36km/h = 10m/s

v = 0m/s

a = -0.5m/s^2

By Newton's First Equation of Motion,

 v = u + a t

0 = 10 - 0.5t

0.5t = 10

t = 10/ 0.5 = 20 sec

The time it will take to come to rest is 20 seconds