Chapter 7, Problem 49RE

a.

To determine

To Find: The velocity as a function of time t

Answer to Problem 49RE

The velocity as a function of time t $\text{ =}2t+3t^2+4$

Explanation of Solution

Given information:

The acceleration of a particle moving along a co-ordinate line is $\frac{d^{2}s}{d{t}^2}=2+6t\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\sec }^2$}\right.$

At $t=0$ the velocity is $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$

Formula used:

Acceleration

  $a=\frac{d^{2}s}{d{t}^2}$

Then, velocity

  $v=\frac{ds}{dt}$

Calculation:

Given that

  $\frac{d^{2}s}{d{t}^2}=2+6t\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\sec }^2$}\right.$ and at $t=0$ the velocity is $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$

For the velocity function

  $\begin{array}{l}a=\frac{d^{2}s}{d{t}^2}=2+6t\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\sec }^2$}\right.\\ \frac{ds}{dt}={\displaystyle \int \left(2+6t\right)}dt\\ \frac{ds}{dt}=2t+3t^2+c\\ \text{at t=0, }\frac{ds}{dt}=4\\ \text{so, c=4}\\ \text{Velocity = }\frac{ds}{dt}\\ \text{Velocity =}2t+3t^2+4\end{array}$

b.

To determine

To Find: how far does the particle move during the first second of its trip

Answer to Problem 49RE

particle move $6m$ in first second

Explanation of Solution

Given information:

The velocity as a function of time t $\text{ =}2t+3t^2+4$

Formula used:

Acceleration

  $a=\frac{d^{2}s}{d{t}^2}$

Then, velocity

  $v=\frac{ds}{dt}$

Calculation:

The velocity as a function of time t $\text{ =}2t+3t^2+4$

  $\begin{array}{l}\frac{ds}{dt}={\displaystyle \int \left(2+6t\right)}dt\\ \frac{ds}{dt}=2t+3t^2+c\\ \text{at t=0, }\frac{ds}{dt}=4\\ \frac{ds}{dt}\\ \frac{ds}{dt}\text{=}2t+3t^2+4\\ s={\displaystyle \underset{0}{\overset{1}{\int }}\left(2t+3t^2+4\right)}dt\\ s={\left[t^2+t^3+4t\right]}_0^1\\ s=6\end{array}$

particle move $6m$ in first second