Chapter 3.1, Problem 3PP

(a)

To determine

To Calculate: The average acceleration ( $a_1$ ) of the train during the given time interval $\left(0.0\text{s to }5.0\mathrm{s}\right)$ using the given graph.

Answer to Problem 3PP

The average acceleration ( $a_1$ ) of the train during the given time interval $\left(0.0\text{s to }5.0\mathrm{s}\right)$ is

  $2.0\text{m}\cdot {\text{s}}^{\text{-2}}$ .

Explanation of Solution

Given:

Time interval for which average acceleration ( $a_1$ ) is to be calculated is $t_i=0.0\text{s}$ to $t_f=5.0\text{s}$ .

Formula used:

  $\text{(}{a}_1\text{)=}\frac{\left[\text{Velocity(}{v}_f\text{)}\text{at}\text{time}{t}_f\right]\text{-}\left[\text{Velocity(}{v}_i\text{)}\text{at}\text{time}{t}_i\right]}{t_f-t_i}$

  $a_1=\frac{v_f-v_i}{t_f-t_i}$

Calculation:

From the graph, it can be seen clearly that

Velocity $(v_f)$ at time $t_f$ = 10 $\text{m}\cdot {\text{s}}^{\text{-1}}$

Velocity $(v_i)$ at time $t_i$ = 0 $\text{m}\cdot {\text{s}}^{\text{-1}}$

Substituting the given values

  $a_1=\frac{10.0-0.0}{5.0-0.0}$

  $a_1=2.0\text{m}\cdot {\text{s}}^{\text{-2}}$

Conclusion:

Thus, the value of the average acceleration ( $a_1$ ) of the train during the given time interval (0.0 s to 5.0 s) is $2.0\text{m}\cdot {\text{s}}^{\text{-2}}$ .

(b)

To determine

The average acceleration ( $a_2$ ) of the train during the given time interval $\left(15.0\text{s to }20.0\text{s}\right)$ using the given graph.

Answer to Problem 3PP

  $\text{-1}\text{.20}\text{m}\cdot {\text{s}}^{\text{-2}}$

Explanation of Solution

Given:

Time interval for which average acceleration ( $a_2$ ) is to be calculated is $t_i=15.0\text{s}$ to $t_f=20.0\text{s}$ .

Formula used:

  $\text{(}{a}_2\text{)=}\frac{\left[\text{Velocity(}{v}_f\text{)}\text{at}\text{time}{t}_f\right]\text{-}\left[\text{Velocity(}{v}_i\text{)}\text{at}\text{time}{t}_i\right]}{t_f-t_i}$

  $a_2=\frac{v_f-v_i}{t_f-t_i}$

Calculation:

From the graph, it can be seen clearly that:

Velocity $(v_f)$ at time $t_f$

  $\text{=4}\text{.0}\text{m}\cdot {\text{s}}^{\text{-1}}$

Velocity $(v_i)$ at time $t_i$ = 10.0 $\text{m}\cdot {\text{s}}^{\text{-1}}$

Substituting the given values:

  $a_2=\frac{4.0-10.0}{20.0-15.0}$

  $a_2=-1.20\text{m}\cdot {\text{s}}^{\text{-2}}$

Conclusion:

Thus, the average acceleration ( $a_2$ ) of the train during the given time interval (15.0 s to 20.0 s) is -1.20 $\text{m}\cdot {\text{s}}^{\text{-2}}$ .

(c)

To determine

To Calculate: The average acceleration ( $a_3$ ) of the train during the given time interval $\left(\text{0}\text{.0 s to 40}\text{.0 s}\right)$ using the given graph.

Answer to Problem 3PP

0.0 $\text{m}\cdot {\text{s}}^{\text{-2}}$

Explanation of Solution

Given:

Time interval for which average acceleration ( $a_3$ ) is to be calculated is $t_i=0.0\text{s}$ to $t_f=40.0\text{s}$ .

Formula used:

  $\text{(}{a}_3\text{)=}\frac{\left[\text{Velocity(}{v}_f\text{)}\text{at}\text{time}{t}_f\right]\text{-}\left[\text{Velocity(}{v}_i\text{)}\text{at}\text{time}{t}_i\right]}{t_f-t_i}$

  $a_3=\frac{v_f-v_i}{t_f-t_i}$

Calculation:

From the graph, it can be seen clearly that;

Velocity $(v_f)$ at time $t_f$

  $\text{=}0.0\text{m}\cdot {\text{s}}^{\text{-1}}$

Velocity $(v_i)$ at time $t_i$

  $\text{=}0.0\text{m}\cdot {\text{s}}^{\text{-1}}$

Substituting the given values

  $a_3=\frac{0.0-0.0}{40.0-0.0}$

  $a_3\text{=}0.0\text{m}\cdot {\text{s}}^{\text{-1}}$

Conclusion:

Thus, the average acceleration ( $a_3$ ) of the train during the given time interval $\left(\text{0}\text{.0 s to 40}\text{.0 s}\right)$ is

  $0.0\text{m}\cdot {\text{s}}^{\text{-1}}$ .