Question

A particle with a velocity of 2ms at t=0 moves along a straight line with a constant acceleration of 0.2 m/S2. Find the displacement of the particle in 10s.

Answer 1

Answer:

• The displacement (S) is 30 m.

Given:

1. Acceleration (a) = 0.2 m/s².
2. Time (t) = 10 secs.

Explanation:

$\rule{300}{1.5}$

As given that at t = 0 sec its velocity is 2 m/s, so Initial velocity (u) is 2 m/s.

From the formula we know,

$\bigstar\;\underline{\boxed{\sf S=ut+\dfrac{1}{2}\;at^{2}}}$

Here,

• S Denotes displacement.
• u Denotes initial velocity.
• t Denotes time.
• a Denotes acceleration.

Substituting the values,

$\longrightarrow\sf S=2\times 10+\dfrac{1}{2}\times 0.2 \times (10)^{2}\\\\\\\longrightarrow\sf S=20+0.1\times (10)^{2}\\\\\\\longrightarrow\sf S=20+0.1\times 100\\\\\\\longrightarrow\sf S=20+10\\\\\\\longrightarrow S=30\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf S=30\;m}}}}$

∴ The displacement (S) is 30 m.

Diagram:

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6) \put(0,0){\line(1,0){4}}\put(4.1,0){ \dots\dots\dots\dots }\put(6,0){\line(1,0){1}}\put(7,0){\circle*{0.15}}\put(0,0){\circle*{0.15}}\put(-0.5,-0.5){ \sf u=2\;m/s }\put(-0.4,0.5){\sf t\;=\;0}\put(1,2){\vector(1,0){3}}\put(4.1,1.9){ \sf a\;=\;0.2\;m/s^{2} }\put(6.6,-0.5){ \sf u=v\;m/s }\put(6.6,0.5){\sf t\;=\;10}\put(4,-1){\vector(-1,0){4}}\put(4.4,-1){\vector(1,0){2.7}}\put(4.15,-1.1){ \sf S }\end{picture}$

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathbb\blue{ANSWER:-}$

The displacement (S) is 30 m.