Question

A particle starts to move to a straight line from a point with velocity 10m/s and acceleration-2m/s². Find the position and velocity of the particle at (i)t= 5s (ii)t= 10s​

Answer 1

Initial Velocity = 10 m/sec

Acceleration = (-2)m/sec²

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From 1st equation of motion :-

$v = u + at$

Hence,

At t= 5 seconds

$v = 10 + ( - 2)5 \\ \\ \\v = 10 - 10 \\ \\ \\v = 0$

Thus,

At t=5 seconds, the velocity is 0 m/sec

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At t= 10 seconds

$v = 10 + ( - 2)10 \\ \\ \\v = 10 -20 \\ \\ \\v = ( - 10)$

At t=10 seconds, the velocity of the body is (-10) m/sec

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From the 3rd equation of motion we have :-

$2as = {v}^{2} - {u}^{2}$

Hence,

At t=5 seconds

$2( - 2)s = 0 - {10}^{2} \\ \\ \\ - 4s = - 100 \\ \\ \\s = \frac{100}{4} \\ \\ \\s = 25 \: \: m$

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At t=10 seconds

$s = ut + \frac{1}{2}a {t}^{2} \\ \\ \\s = 10 \times 10 + \frac{1}{2}( - 2) \times {10}^{2} \\ \\ \\s = 100 - 100 \\ \\ \\s = 0 \: \: m$

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Answer 2

Answer:

Initial Velocity = 10 m/sec

Acceleration = (-2)m/sec²

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From 1st equation of motion :-

\begin{gathered}v = u + at \\ \\ \\\end{gathered}

v=u+at

Hence,

At t= 5 seconds

\begin{gathered}v = 10 + ( - 2)5 \\ \\ \\v = 10 - 10 \\ \\ \\v = 0\end{gathered}

v=10+(−2)5

v=10−10

v=0

Thus,

At t=5 seconds, the velocity is 0 m/sec

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At t= 10 seconds

\begin{gathered}v = 10 + ( - 2)10 \\ \\ \\v = 10 -20 \\ \\ \\v = ( - 10)\end{gathered}

v=10+(−2)10

v=10−20

v=(−10)

At t=10 seconds, the velocity of the body is (-10) m/sec

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From the 3rd equation of motion we have :-

\begin{gathered}2as = {v}^{2} - {u}^{2} \\ \\ \\\end{gathered}

2as=v

2

−u

2

Hence,

At t=5 seconds

\begin{gathered}2( - 2)s = 0 - {10}^{2} \\ \\ \\ - 4s = - 100 \\ \\ \\s = \frac{100}{4} \\ \\ \\s = 25 \: \: m \end{gathered}

2(−2)s=0−10

2

−4s=−100

s=

4

100

s=25m

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At t=10 seconds

\begin{gathered}s = ut + \frac{1}{2}a {t}^{2} \\ \\ \\s = 10 \times 10 + \frac{1}{2}( - 2) \times {10}^{2} \\ \\ \\s = 100 - 100 \\ \\ \\s = 0 \: \: m\end{gathered}

s=ut+

2

1

at

2

s=10×10+

2

1

(−2)×10

2

s=100−100

s=0m

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