KCET · Physics · Motion In One Dimension
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time \(t\) is proportional to
- A \(t^{1 / 2}\)
- B \(t\)
- C \(t^{3 / 2}\)
- D \(t^{2}\)
Answer & Solution
Correct Answer
(B) \(t\)
Step-by-step Solution
Detailed explanation
Let in time \(t\), the body will acquire a velocity of \(v\) from rest.
As, power \((P)=\) rate of doing work \(=\frac{d W}{d t}\)
\(=\mathbf{F} \cdot \frac{d \mathbf{x}}{d t}=\mathbf{F} \cdot \mathbf{v}\)
For motion in one dimension,
\(P=F v \cos 0^{\circ}=F v...(i)\)
From first equation of motion,
\(\begin{aligned}
&v=u+a t \\
&v=a t ...(ii) (\because u=0)
\end{aligned}\)
From Eqs. (i) and (ii), we get
\(\begin{aligned}
P &=F \times a t=m a \times a t \quad(\because F=m a) \\
&=m a^{2} t
\end{aligned}\)
As \(m a^{2}\) is constant.
So, \(\quad P \propto t\)
As, power \((P)=\) rate of doing work \(=\frac{d W}{d t}\)
\(=\mathbf{F} \cdot \frac{d \mathbf{x}}{d t}=\mathbf{F} \cdot \mathbf{v}\)
For motion in one dimension,
\(P=F v \cos 0^{\circ}=F v...(i)\)
From first equation of motion,
\(\begin{aligned}
&v=u+a t \\
&v=a t ...(ii) (\because u=0)
\end{aligned}\)
From Eqs. (i) and (ii), we get
\(\begin{aligned}
P &=F \times a t=m a \times a t \quad(\because F=m a) \\
&=m a^{2} t
\end{aligned}\)
As \(m a^{2}\) is constant.
So, \(\quad P \propto t\)
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