Chapter 2, Problem 11P

A particle starts from rest and accelerates as shown in Figure P2.11. Determine (a) the particle’s speed at t = 10.0 s and at t = 20.0 s, and (b) the distance traveled in the first 20.0 s.

Figure P2.11

To determine

To calculate: The particle’s speed at $t=10.0\text{s}$ and at $t=20.0\text{s}$.

Answer to Problem 11P

The particle’s speed at $t=10.0\text{s}$ is $20\text{m}/\text{s}$ and at $t=20.0\text{s}$ is $5\text{m}/\text{s}$.

Explanation of Solution

Section 1:

To calculate: The particle’s speed at $t=10.0\text{s}$.

Answer: The particle’s speed at $t=10.0\text{s}$ is $20\text{m}/\text{s}$.

The following graph shows the graph of acceleration versus time for the particle.

Figure I

Formula to calculate the speed of the particle at a given time instant is,

$v_f=v_i+at$

• $v_f$ is the final speed.
• $v_i$ is the initial speed.
• $a$ is the acceleration of the particle at that time instant.
• $t$ is the time period.

Form the shown graph in Figure I, the initial velocity of the particle zero. Similarly, the acceleration of the particle at $t=10.0\text{s}$ is $2\text{m}/{\text{s}}^2$, and the time period from $0\text{s}$ to $10\text{s}$ is $10\text{s}$.

Substitute $0\text{m}/\text{s}$ for $v_i$, $2\text{m}/{\text{s}}^2$ for $a$ and $10\text{s}$ for $t$ in the above equation to find $v_f$.

$\begin{array}{c}v=\left(0\text{m}/\text{s}\right)+\left(2\text{m}/{\text{s}}^2\right)\left(10\text{s}\right)\\ =20\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the particle’s speed at $t=10.0\text{s}$ is $20\text{m}/\text{s}$.

Section 2:

To calculate: The particle’s speed at $t=20.0\text{s}$.

Answer: The particle’s speed at $t=20.0\text{s}$ is $5\text{m}/\text{s}$.

Formula to calculate the speed of the particle at a given time instant is,

$v_{20\text{s}}=u_{15\text{s}}+\left(a_{15\text{s}-20\text{}\text{s}}\right)t_{\left(15\text{s}-20\text{s}\right)}$

• $v_{20\text{s}}$ is the speed of the particle at $t=20\text{}\text{s}$.
• $u_{15\text{s}}$ is the speed of the particle at $t=15\text{s}$.
• $a_{15\text{s}-20\text{}\text{s}}$ is the acceleration of the particle during the time instant $t=15\text{s}$ to $t=20\text{}\text{s}$.
• $t_{\left(15\text{s}-20\text{s}\right)}$ is the time period form $t=15\text{s}$ to $t=20\text{}\text{s}$.

Form the shown graph in Figure I, the acceleration of the particle from $t=10.0\text{s}$ to $t=15\text{s}$ is zero. Thus, the velocity of the particle at $t=15\text{s}$ is equal to the velocity at $t=10.0\text{s}$, that is $20\text{m}/\text{s}$. Similarly, the acceleration of the particle from $t=15\text{s}$ to $t=20\text{}\text{s}$ is $-3\text{m}/{\text{s}}^2$ and the time duration is $5\text{s}$.

Substitute $20\text{m}/\text{s}$ for $u_{15\text{s}}$, $-3\text{m}/{\text{s}}^2$ for $a_{15\text{s}-20\text{}\text{s}}$ and $5\text{s}$ for $t_{\left(15\text{s}-20\text{s}\right)}$ in the above equation to find $v_{20\text{s}}$.

$\begin{array}{c}{v}_{20\text{s}}=\left(20\text{m}/\text{s}\right)+\left(-3\text{m}/{\text{s}}^2\right)\left(5\text{s}\right)\\ =\left(20\text{m}/\text{s}\right)-\left(15\text{m}/\text{s}\right)\\ =5\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the particle’s speed at $t=20.0\text{s}$ is $5\text{m}/\text{s}$.

To determine

To calculate: The distance travelled in first $20\text{s}$.

Answer to Problem 11P

The distance travelled in the first $20\text{s}$ is $263\text{m}$.

Explanation of Solution

Given information:

The particle starts from rest and accelerates as shown in Figure-(I).

Formula to calculate the distance travelled by the particle in first $10\text{s}$ is,

$s_{10\text{s}}=u_{0\text{s}}{t}_{10\text{s}}+\frac{1}{2}{a}_{10\text{s}}{t}_{10\text{s}}{}^2$

• $s_{10\text{s}}$ is the distance travelled first $10\text{s}$.
• $u_{0\text{s}}$ is the initial velocity of the particle $t=0\text{s}$.
• $t_{10\text{s}}$ is the time duration from $0\text{s}$ to $10\text{s}$.
• $a_{10\text{s}}$ is the acceleration of the particle from $0\text{s}$ to $10\text{s}$.

Formula to calculate the distance travelled by the particle form $10\text{s}$ to $15\text{s}$ is,

$s_{15\text{s}}=u_{10\text{s}}{t}_{15\text{s}}+\frac{1}{2}{a}_{\left(10\text{s-15}\text{s}\right)}{t}_{15\text{s}}{}^2$

• $s_{15\text{s}}$ is the distance travelled form $10\text{s}$ to $15\text{s}$.
• $u_{10\text{s}}$ is the velocity of the particle at $t=10\text{s}$.
• $t_{15\text{s}}$ is the time duration form $10\text{s}$ to $15\text{s}$.
• $a_{\left(10\text{s-15}\text{s}\right)}$ is the acceleration of the particle from $10\text{s}$ to $15\text{s}$.

Formula to calculate the distance travelled by the particle form $15\text{s}$ to $20\text{s}$ is,

$s_{20\text{s}}=u_{15\text{s}}{t}_{20\text{s}}+\frac{1}{2}{a}_{\left(15\text{s-20}\text{s}\right)}{t}_{20\text{s}}{}^2$

• $s_{20\text{s}}$ is the distance travelled form $15\text{s}$ to $20\text{s}$.
• $u_{15\text{s}}$ is the velocity of the particle at $t=15\text{s}$.
• $t_{20\text{s}}$ is the time duration form $15\text{s}$ to $20\text{s}$.
• $a_{\left(15\text{s-20}\text{s}\right)}$ is the acceleration of the particle from $15\text{s}$ to $20\text{s}$.

Formula to calculate the total distance travelled by the particle from starting to the end is,

$s_{\text{total}}=s_{10\text{s}}+s_{15\text{s}}+s_{20\text{s}}$

• $s_{\text{total}}$ is the total distance traveled y the particle.

Substitute $u_{0\text{s}}{t}_{10\text{s}}+\frac{1}{2}{a}_{10\text{s}}{t}_{10\text{s}}{}^2$ for $s_{10\text{s}}$, $u_{10\text{s}}{t}_{15\text{s}}+\frac{1}{2}{a}_{\left(10\text{s-15}\text{s}\right)}{t}_{15\text{s}}{}^2$ for $s_{15\text{s}}$ and $u_{15\text{s}}{t}_{20\text{s}}+\frac{1}{2}{a}_{\left(15\text{s-20}\text{s}\right)}{t}_{20\text{s}}{}^2$ for $s_{20\text{s}}$ in the above equation to find $s_{\text{total}}$.

$s_{\text{total}}=\left(u_{0\text{s}}{t}_{10\text{s}}+\frac{1}{2}{a}_{10\text{s}}{t}_{10\text{s}}{}^2\right)+\left(u_{10\text{s}}{t}_{15\text{s}}+\frac{1}{2}{a}_{\left(10\text{s-15}\text{s}\right)}{t}_{15\text{s}}{}^2\right)+\left(u_{15\text{s}}{t}_{20\text{s}}+\frac{1}{2}{a}_{\left(15\text{s-20}\text{s}\right)}{t}_{20\text{s}}{}^2\right)$

Substitute $0\text{m}/\text{s}$ for $u_{0\text{s}}$, $10\text{s}$ for $t_{10\text{s}}$, $2\text{m}/{\text{s}}^2$ for $a_{10\text{s}}$, $20\text{m}/\text{s}$ for $u_{10\text{s}}$, $5\text{s}$ for $t_{15\text{s}}$, $0\text{m}/{\text{s}}^2$ for $a_{\left(10\text{s-15}\text{s}\right)}$, $20\text{m}/\text{s}$ for $u_{15\text{s}}$, $5\text{s}$ for $t_{20\text{s}}$ and $-3\text{m}/{\text{s}}^2$ for $a_{15\text{s}-20\text{}\text{s}}$ in the above equation to find $s_{\text{total}}$.

$\begin{array}{c}{s}_{\text{total}}=\left[\begin{array}{l}\left(\left(0\text{m}/\text{s}\right)\left(10\text{s}\right)+\frac{1}{2}\left(2\text{m}/{\text{s}}^2\right){\left(\left(10\text{s}\right)\right)}^2\right)+\left(\left(20\text{m}/\text{s}\right)\left(5\text{s}\right)+\frac{1}{2}\left(0\text{m}/{\text{s}}^2\right){\left(\left(5\text{s}\right)\right)}^2\right)\\ +\left(\left(20\text{m}/\text{s}\right)\left(\left(5\text{s}\right)\right)+\frac{1}{2}\left(-3\text{m}/{\text{s}}^2\right){\left(\left(5\text{s}\right)\right)}^2\right)\end{array}\right]\\ =\left(100\text{m}\right)+\left(100\text{m}\right)+\left(\left(100\text{m}\right)-\left(37.5\text{m}\right)\right)\\ =262.5\text{m}\\ =263\text{m}\end{array}$

Thus, the distance travelled in the first $20\text{s}$ is $263\text{m}$.

Conclusion:

Therefore, the distance travelled in the first $20\text{s}$ is $263\text{m}$.