Chapter 2, Problem 55QAP

To determine

(a)

The magnitude of constant acceleration for the given interval of time.

Answer to Problem 55QAP

The magnitude of acceleration for the first five second is $0.5556{\text{ms}}^{\text{-2}}$. The magnitude of acceleration for the next five second is $0.5566{\text{ms}}^{\text{-2}}$.

Explanation of Solution

Given info:

• The initial speed speed and final speed of the car is $35{\text{kmh}}^{\text{-1}}$and $45{\text{kmh}}^{\text{-1}}$for the first $5\text{s}$.
• Convert kilometer per hour to meter per second.

  $\begin{array}{c}35\text{km/h}\text{=}\text{35}\frac{\text{km}}{\text{h}}\text{×}\frac{\text{1000}\text{m}}{\text{1}\text{km}}\text{×}\frac{\text{1}\text{h}}{\text{3600}\text{s}}\\ =9.722\text{m/s}\\ 45\text{km/h}\text{=}45\frac{\text{km}}{\text{h}}\text{×}\frac{\text{1000}\text{m}}{\text{1}\text{km}}\text{×}\frac{\text{1}\text{h}}{\text{3600}\text{s}}\\ =12.5\text{m/s}\end{array}$

• The initial speed speed and final speed of the car is $65{\text{kmh}}^{\text{-1}}$and $75{\text{kmh}}^{\text{-1}}$for the next $5\text{s}$.

Convert kilometer per hour to meter per second.

  $\begin{array}{c}65\text{km/h}\text{=}\text{65}\frac{\text{km}}{\text{h}}\text{×}\frac{\text{1000}\text{m}}{\text{1}\text{km}}\text{×}\frac{\text{1}\text{h}}{\text{3600}\text{s}}\\ =18.05\text{m/s}\\ 75\text{km/h}\text{=}75\frac{\text{km}}{\text{h}}\text{×}\frac{\text{1000}\text{m}}{\text{1}\text{km}}\text{×}\frac{\text{1}\text{h}}{\text{3600}\text{s}}\\ =20.833\text{m/s}\end{array}$

Formula used:

Use first equation of motion to find the acceleration.

  $\begin{array}{l}v=u+at\\ a=\frac{v-u}{t}\text{Equation-1}\end{array}$

Here, $v$ is the finalvelocity $u$ is the initial velocity and $t$ is the time interval.

Calculation:

Finding the acceleration:

For first $5\text{s}$ :

Substituting the all known values in equation-1 to find acceleration,

  $\begin{array}{c}{a}_1=\frac{{\left(12.5{\text{ms}}^{\text{-1}}\right)}^2-{\left(9.722{\text{ms}}^{\text{-1}}\right)}^2}{5\text{s}}\\ =0.5556{\text{ms}}^{\text{-2}}\text{Equation-2}\\ \end{array}$

For Last $5\text{s}$ :

Substituting the all known values in equation-1 to find acceleration,

  $\begin{array}{c}{a}_2=\frac{{\left(20.833{\text{ms}}^{\text{-1}}\right)}^2-{\left(18.05{\text{ms}}^{\text{-1}}\right)}^2}{5\text{s}}\\ =0.5566{\text{ms}}^{\text{-2}}\text{Equation-3}\end{array}$

Conclusion:

From equation-3 and 4 magnitude of acceleration for the first five second is $0.5556{\text{ms}}^{\text{-2}}$. The magnitude of acceleration for the next five second is $0.5566{\text{ms}}^{\text{-2}}$.

To determine

(b)

The magnitude of constant acceleration for the given interval of time.

Answer to Problem 55QAP

The magnitude of acceleration for the first five second is $0.5556{\text{ms}}^{\text{-2}}$. The magnitude of acceleration for the next five second is $0.5566{\text{ms}}^{\text{-2}}$.

Explanation of Solution

Given info:

• The initial speed speed and final speed of the car is $9.722{\text{ms}}^{\text{-1}}$and $12.5{\text{ms}}^{\text{-1}}$for the first $5\text{s}$.
• The initial speed speed and final speed of the car is $18.05{\text{ms}}^{\text{-1}}$and $20.833{\text{ms}}^{\text{-1}}$for the next $5\text{s}$.
• Magnitude of acceleration for the first five second is $0.5556{\text{ms}}^{\text{-2}}$. The magnitude of acceleration for the next five second is $0.5566{\text{ms}}^{\text{-2}}$.

Formula used:

Use third equation of motion to find the distance travelled.

  $\begin{array}{l}{v}^2=u^2+2a\text{s}\\ s=\frac{v^2-u^2}{2\text{a}}\text{Equation-4}\end{array}$

Here, $v$ is the finalvelocity $u$ is the initial velocity and $\text{a}$ is the acceleration.

Calculation:

Finding the distance travelled:

For first $5\text{s}$ :

Substituting the all known values in equation-2 to find distance,

  $\begin{array}{c}{s}_1=\frac{{\left(12.5{\text{ms}}^{\text{-1}}\right)}^2-{\left(9.722{\text{ms}}^{\text{-1}}\right)}^2}{2\times 0.5556{\text{ms}}^{-2}}\\ =55.5\text{m}\text{Equation-5}\\ s_1=\frac{{\left(20.833{\text{ms}}^{\text{-1}}\right)}^2-{\left(18.05{\text{ms}}^{\text{-1}}\right)}^2}{2\times 0.5566{\text{ms}}^{-2}}\\ =97.2\text{m}\text{Equation-6}\end{array}$

For Last $5\text{s}$ :

Substituting the all known values in equation-1 to find acceleration,

Conclusion:

From equation-5 and 6distance travelledin the first five second is $55.5\text{m}$. The magnitude of acceleration for the next five second is $97.2\text{m}$.