Question

A constant retarding force of 5ON is applied to a body of mass 30kg moving initially with a speed of 18m/s. How long does the body take to come to halt?

Answer 1

◑ Given ◑

A constant retarding force of 5ON is applied to a body of mass 30kg moving initially with a speed of 18m/s.

◑ To find ◑

• How long does the body take to come to halt?

◑ Solution ◑

$\\\bold\dag\:{\sf{\red{Retarding\:Force=50N}}}$

$\bold\dag\:{\sf{\red{Mass\:of\:body=30kg}}}$

$\bold\dag\:{\sf{\red{Initial\:velocity=18m/s}}}$

$\bold\dag\:{\sf{\red{Final\:velocity=0}}}$

$\qquad{\underline{\mathfrak{\qquad As\:we\:know\:that\qquad}}}$

$\qquad{\large\bullet}\:{\bf{\blue{Force=Mass\times Accleration}}}$

$\implies\sf F = m\times a$

$\implies\sf -50=30 \times a \quad \bf (Retarding\:force)$

$\implies\sf -50=30a$

$\implies\sf a = \cancel\dfrac{(-50)}{30}$

$\implies\sf a = \dfrac{-5}{3} m/s^2$

$\therefore{\underline{\boxed{\bf{Acceleration\:of\:a\:body\:is\:\dfrac{-5}{3}m/s^2}}}}$

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Now we have to find out time taken to come to halt

$\\\qquad\large\bullet\:{\bf{\blue{v=u+at}}}$

Where,

• v = Final velocity
• u = initial velocity
• t = time
• a = accleration

$\\\qquad \quad$ Substitute the values

$\\\implies\sf v = u + at$

$\implies\sf 0 = 18 + \dfrac{(-5)}{3}\times t$

$\implies\sf - 18 = \dfrac{-5t}{3}$

$\implies\sf - 18 \times 3 = - 5t$

$\implies\sf t = \dfrac{18\times 3}{5}$

$\implies\sf t = 10.8 s$

$\therefore{\underline{\boxed{\bf{Time\:taken \: to\: come \: to \: halt \: is \: 10.8 s}}}}$

$\:$____________________________________________

Answer 2

Time taken to come to halt is 10.8 sec