Question

A ball is initially moving with a velocity 0.5 m/s. Its velocity decreases at a rate of

0.05 m/s2.

(a) How much time will it take to stop?​

Answer 1

Given :

• Initial velocity of the ball is 0.5 m/s
• Rate of decreasing of velocity = 0.05 m/s²

To find :

• The time taken by the ball to come to rest

Solution :

The Final velocity of the ball is Zero. [ Since it comes to rest]

We are given that the velocity is decreasing at a rate of 0.05 m/s² which is nothing but the deceleration of particle(decreasing acceleration) . Then acceleration of the ball is -0.05 m/s² [ negative sign indicates that it is decreasing]

We have ,

• v = 0 m/s
• u = 0.5 m/s
• a = -0.05 m/s²
• t = ?

From first equation of motion ,

$\large\star \: { \boxed{ \sf{ \purple{v = u + at}}}}$

Where ,

• v is final velocity
• u is initial velocity
• a is acceleration
• t is time

By substituting the values ,

$: \implies \sf \: 0 = 0.5 \: m {s}^{ - 1} + ( - 0.05 \: m {s}^{ - 2} )(t) \\ \\ : \implies \sf \: 0 = 0.5 \: m {s}^{ - 1} + - 0.05t \: m {s}^{ - 2} \\ \\ \ \: : \implies \sf \: - 0.5 \: m {s}^{ - 1} = - 0.05t \: m {s}^{ - 2} \\ \\ : \implies \sf \not{ - } \: 0.5 \: m {s}^{ - 1} = \not{ - } \: 0.05t \: m {s}^{ - 2} \\ \\ : \implies \sf \: 0.5 \: m {s}^{ - 1} = 0.05t \: m {s}^{ - 2} \\ \\ : \implies \sf \: t = \frac{0.5 \: m {s}^{ - 1} }{0.05 \: m {s}^{ - 2} } \\ \\ : \implies \sf t = 10 \: s$

Hence , The time taken by the ball to come to rest is 10 sec

Answer 2

Answer:

Given :

Initial velocity of the ball is 0.5 m/s

Rate of decreasing of velocity = 0.05 m/s²

To find :

The time taken by the ball to come to rest

Solution :

The Final velocity of the ball is Zero. [ Since it comes to rest]

We are given that the velocity is decreasing at a rate of 0.05 m/s² which is nothing but the deceleration of particle(decreasing acceleration) . Then acceleration of the ball is -0.05 m/s² [ negative sign indicates that it is decreasing]

We have ,

v = 0 m/s

u = 0.5 m/s

a = -0.05 m/s²

t = ?

From first equation of motion ,

\large\star \: { \boxed{ \sf{ \purple{v = u + at}}}}⋆

v=u+at

Where ,

v is final velocity

u is initial velocity

a is acceleration

t is time

By substituting the values ,

\begin{gathered} : \implies \sf \: 0 = 0.5 \: m {s}^{ - 1} + ( - 0.05 \: m {s}^{ - 2} )(t) \\ \\ : \implies \sf \: 0 = 0.5 \: m {s}^{ - 1} + - 0.05t \: m {s}^{ - 2} \\ \\ \ \: : \implies \sf \: - 0.5 \: m {s}^{ - 1} = - 0.05t \: m {s}^{ - 2} \\ \\ : \implies \sf \not{ - } \: 0.5 \: m {s}^{ - 1} = \not{ - } \: 0.05t \: m {s}^{ - 2} \\ \\ : \implies \sf \: 0.5 \: m {s}^{ - 1} = 0.05t \: m {s}^{ - 2} \\ \\ : \implies \sf \: t = \frac{0.5 \: m {s}^{ - 1} }{0.05 \: m {s}^{ - 2} } \\ \\ : \implies \sf t = 10 \: s\end{gathered}

:⟹0=0.5ms

−1

+(−0.05ms

−2

)(t)

:⟹0=0.5ms

−1

+−0.05tms

−2

:⟹−0.5ms

−1

=−0.05tms

−2

:⟹



−0.5ms

−1

=



−0.05tms

−2

:⟹0.5ms

−1

=0.05tms

−2

:⟹t=

0.05ms

−2

0.5ms

−1

:⟹t=10s

Hence , The time taken by the ball to come to rest is 10 sec