a particle starts to move in a straight line from a point with velocity 10m/s and acceleration-2m/s2 . find the position and velocity of the particle at (i)5s (ii)10s
Answers
Answer:
Initial Velocity = 10 m/sec
Acceleration = (-2)m/sec²
$$\rule{200}{2}$$
From 1st equation of motion :-
$$\begin{gathered}v = u + at \\ \\ \\\end{gathered}$$
Hence,
At t= 5 seconds
$$\begin{gathered}v = 10 + ( - 2)5 \\ \\ \\v = 10 - 10 \\ \\ \\v = 0\end{gathered}$$
Thus,
At t=5 seconds, the velocity is 0 m/sec
$$\rule{200}{2}$$
At t= 10 seconds
$$\begin{gathered}v = 10 + ( - 2)10 \\ \\ \\v = 10 -20 \\ \\ \\v = ( - 10)\end{gathered}$$
At t=10 seconds, the velocity of the body is (-10) m/sec
$$\rule{200}{2}$$
From the 3rd equation of motion we have :-
$$\begin{gathered}2as = {v}^{2} - {u}^{2} \\ \\ \\\end{gathered}$$
Hence,
At t=5 seconds
$$\begin{gathered}2( - 2)s = 0 - {10}^{2} \\ \\ \\ - 4s = - 100 \\ \\ \\s = \frac{100}{4} \\ \\ \\s = 25 \: \: m\end{gathered}$$
$$\rule{200}{2}$$
At t=10 seconds
$$\begin{gathered}s = ut + \frac{1}{2}a {t}^{2} \\ \\ \\s = 10 \times 10 + \frac{1}{2}( - 2) \times {10}^{2} \\ \\ \\s = 100 - 100 \\ \\ \\s = 0 \: \: m\end{gathered}$$
$$\rule{200}{2}$$
Initial Velocity = 10 m/sec
Acceleration = (-2)m/sec²
From 1st equation of motion :-
Hence,
At t= 5 seconds
Thus,
At t=5 seconds, the velocity is 0 m/sec
At t= 10 seconds
At t=10 seconds, the velocity of the body is (-10) m/sec
From the 3rd equation of motion we have :-
Hence,
At t=5 seconds
At t=10 second
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