Q81P - Fundamentals Of Physics

Question

A particle’s acceleration along an x axis is $\mathit{a}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{t}$, with t in seconds and a in meters per second squared. At $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$, its velocity is $\mathbf{+}\mathbf{17}\mathbf{ }\mathit{m}\mathbf{/}\mathit{s}$. What is its velocity at $\mathit{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$?

Step-by-Step Solution

Answer

The velocity of the particle at $t=4.0 s \mathrm{is} 47 m/s$.

Step 1: Given information

$$\begin{gathered} a=5.0t \\ t=2.0 \mathrm{s} \\ v=+17 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

Step 2: To understand the concept and relation between displacement, velocity and acceleration

If we are given with the displacement equation with respect to time and we differentiate the equation with respect to time, resulted equation will be velocity equation. If we differentiate velocity equation with respect to time, resulted equation will be acceleration equation. 

If we are given with the acceleration equation with respect to time and we integrate with respect to time, resulted equation will be velocity equation. If we integrate velocity equation with respect to time, resulted equation will be displacement equation.

 

Formula:

$v=\int a \mathrm{d}t$

Step 3: Calculation for the velocity at the given time.

The velocity of the particle $=\int a \mathrm{d}t$

The velocity of the particle $=\int 5t \mathrm{d}t$

The velocity of the particle at $\left(t=2.0 \sec \right)=5\frac{t^2}{2}+c$

$$\begin{gathered} 17=\frac{5{\left(2\right)}^2}{2}+c \\ 17=10+c \\ C=7 \end{gathered}$$

Using this constant of integration, we can find the velocity of the particle at 4.0 sec as,

The velocity of the particle at $t=4.0 \sec =5\frac{4^2}{2}+\mathrm{c}$

The velocity of the particle at $t=4.0 \sec =5\frac{4^2}{2}+7$

The velocity of the particle at $t=4.0 \sec =47 \mathrm{m}/\mathrm{s}$

Hence, the velocity of the particle is $47 \mathrm{m}/\mathrm{s}$