Question

If body moving with a velocity of 3ms-1 is brought to rest by a constant force after travelling 15m calculate the retardation

Answer 1

$\rule{300}{3}$

❁GIVEN :-

A body moving with a velocity of 3 m/s is brought to rest by a constant force after travelling 15 m.

✪  Initial velocity(u) = 3 m/s

✪ Final velocity(v) = 0 m/s     【as it is brought to rest at last】

✪ Distance travelled(s) = 15 m

❁TO FIND :-

The retardation(deceleration)* of the body.

❁FORMULA TO BE USED :-

v² - u² = 2as

❁SOLUTION :-

Putting the known values in the formula, we get :

(0)² - (3)² = 2 × a × 15

⇒0 = 30a + 9

⇒30a = -9

⇒a = -9/30

⇒a = -3/10

⇒a = -0.3 m/s²

✪ So, the retardation of the body is -0.3 m/s².

$\rule{300}{3}$

*Retardation(or deceleration) always occurs when the velocity of the body is decreased. i.e. Initial velocity(u) > Final velocity(v). Its magnitude is always negative.

Answer 2

Answer :

The retardation is -0.3m/s²

Given :

• A body is moving with a velocity of 3m/s
• It is brought to rest at a constant force after it travelled a distance of 15m

To Find :

• The retardation of the body

Formula to be Used :

$\bullet \: \: \bf {v}^{2} - {u}^{2} = 2as$

Solition :

We are given ,

initial velocity , u = 3m/s

final velocity , v = 0m/s [since the body comes to rest]

and distance , s = 15m

Using equation of motion in the given data :

$\sf \implies {0}^{2} - {3}^{2} = 2 \times a \times 15 \\ \\ \sf \implies - 9 = 30a \\ \\ \sf \implies a = \dfrac{ - 9}{30} \\ \\ \sf \implies a = - \dfrac{ 3}{10} \\ \\ \bf \implies a = - 0.3 \: m {s}^{ - 2}$

The acceleration is -3m/s² . The negative sign signifies that it is retarded.

Thus , the retardation is -0.3m/s²

Other two equations of motion :

$\bullet \bf \: \: \: v = u + at$

$\bf \bullet \: \: \: s = ut + \dfrac{1}{2} at {}^{2}$