Chapter 2, Problem 20P

A particle starts from rest and accelerates as shown in Figure P2.20. 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 P2.20

To determine

The speed of particle at $10.0\text{s}$ and $20.0\text{s}$.

Answer to Problem 20P

The speed of particle at $10.0\text{s}$ and $20.0\text{s}$ are $20.0\text{m/s}$ and $5.00\text{m/s}$.

Explanation of Solution

Given Info: The acceleration till $10.0\text{s}$ is $2.00{\text{m/s}}^{\text{2}}$, the initial velocity is $0$, the time is $10.0\text{s}$, the acceleration in the interval $10.0\text{s}$ to $15.0\text{s}$ is $0$, and the length of the interval is $5.00\text{s}$.

Explanation:

The formula used to calculate the velocity is,

$v=v_0+at$

• $v$ is the final velocity
• $v_0$ is the initial velocity
• $a$ is the acceleration of the particle
• $t$ is the time

Substitute $0$ for $v_0$, $2.00{\text{m/s}}^2$ for $a$ and $\text{10}\text{.0}\text{s}$ for $t$ to find $v$.

$\begin{array}{c}v=0+\left(2{\text{m/s}}^2\right)\left(10.0\text{s}\right)\\ =20.0\text{m/s}\end{array}$

Thus, the speed of particle at $10.0\text{s}$ is $20.0\text{m/s}$.

The formula used to calculate the velocity is,

$v_{15}=v+a_1\Delta t$

• $v$ is the velocity at $10.0\text{s}$
• $v_{15}$ is the velocity at $15.0\text{s}$
• $a_1$ is the acceleration of the particle
• $\Delta t$ is the time

Substitute $20.0\text{m/s}$ for $v$, $0$ for $a_1$ and $\text{5}\text{.00}\text{s}$ for $\Delta t$ to find $v_{15}$.

$\begin{array}{c}{v}_{15}=20.0\text{m/s}+\left(0\right)\left(5.00\text{s}\right)\\ =20.0\text{m/s}\end{array}$

Thus, the speed of particle at $15.0\text{s}$ is $20.0\text{m/s}$.

The acceleration in the interval $15.0\text{s}$ to $20.0\text{s}$ is $-3.00{\text{m/s}}^2$.

The length of the interval is $5.00\text{s}$.

The formula used to calculate the velocity is,

$v_{20}=v_{15}+a_2\Delta t\text{'}$

• $v_{20}$ is the velocity at $20.0\text{s}$
• $v_{15}$ is the velocity at $15.0\text{s}$
• $a_2$ is the acceleration of the particle
• $\Delta t\text{'}$ is the time

Substitute $20.0\text{m/s}$ for $v_{15}$, $-3.00{\text{m/s}}^2$ for $a_2$ and $\text{5}\text{.00}\text{s}$ for $\Delta t\text{'}$ to find $v_{20}$.

$\begin{array}{c}{v}_{20}=20.0\text{m/s}+\left(-3{\text{m/s}}^2\right)\left(5.00\text{s}\right)\\ =5.00\text{m/s}\end{array}$

Thus, the speed of particle at $20.0\text{s}$ is $5.00\text{m/s}$.

Conclusion:

The speed of particle at $10.0\text{s}$ and $20.0\text{s}$ are $20.0\text{m/s}$ and $5.00\text{m/s}$.

To determine

The distance travelled in first $20.0\text{s}$.

Answer to Problem 20P

The distance travelled in first $20.0\text{s}$ is $263\text{m}$.

Explanation of Solution

Given Info:

The acceleration till $10.0\text{s}$ is $2.00{\text{m/s}}^2$.

The initial velocity is $0$.

The time is $10.0\text{s}$.

Explanation:

The formula used to calculate the distance is,

$x=v_{0}t+\frac{1}{2}a{t}^2$

• $x$ is the distance covered
• $v_0$ is the initial velocity
• $a$ is the acceleration of the particle
• $t$ is the time

Substitute $0$ for $v_0$, $2.00{\text{m/s}}^2$ for $a$ and $\text{10}\text{.0}\text{s}$ for $t$ to find $x$.

$\begin{array}{c}x=0\left(10.0\text{s}\right)+\frac{1}{2}\left(2{\text{m/s}}^2\right){\left(10.0\text{s}\right)}^2\\ =100\text{m}\end{array}$

Thus, the distance covered by the particle in $10.0\text{s}$ is $100\text{m}$.

The velocity in the interval $10.0\text{s}$ to $15.0\text{s}$ is $20.0\text{m/s}$.

The distance covered by the particle in $10.0\text{s}$ is $100\text{m}$.

The length of the interval is $5.00\text{s}$.

The formula used to calculate the distance is,

$x_{15}=x+v\Delta t$

• $x$ is the distance covered in $10.0\text{s}$
• $x_{15}$ is the distance covered in $15.0\text{s}$
• $v$ is the speed of the particle
• $\Delta t$ is the time

Substitute $20.0\text{m/s}$ for $v$, $100\text{m}$ for $x$ and $\text{5}\text{.00}\text{s}$ for $\Delta t$ to find $x_{15}$.

$\begin{array}{c}{x}_{15}=100\text{m/s}+\left(20.0\text{m/s}\right)\left(5.00\text{s}\right)\\ =200\text{m/s}\end{array}$

Thus, the distance covered by the particle in $15.0\text{s}$ is $200\text{m}$.

The velocity at $15.0\text{s}$ is $20.0\text{m/s}$.

The distance covered by the particle in $15.0\text{s}$ is $200\text{m}$.

The length of the interval is $5.00\text{s}$.

The formula used to calculate the velocity is,

$x_{20}=x_{15}+v_{15}\Delta t\text{'}+\frac{1}{2}{a}_2{\left(\Delta t\text{'}\right)}^2$

• $x_{20}$ is the distance covered in $20.0\text{s}$
• $x_{15}$ is the distance covered in $15.0\text{s}$
• $v_{15}$ is the speed of the particle at $20.0\text{m/s}$
• $a_2$ is the acceleration of the particle
• $\Delta t\text{'}$ is the time

Substitute $20.0\text{m/s}$ for $v_{15}$, $200\text{m}$ for $x_{15}$ , $-3{\text{m/s}}^2$ for $a_2$ and $\text{5}\text{.00}\text{s}$ for $\Delta t\text{'}$ to find $x_{20}$.

$\begin{array}{c}{x}_{15}=200\text{m/s}+\left(20.0\text{m/s}\right)\left(5.00\text{s}\right)+\frac{1}{2}\left(-3.00{\text{m/s}}^2\right){\left(5.00\text{s}\right)}^2\\ =263\text{m/s}\end{array}$

Thus, the distance covered by the particle in $20.0\text{s}$ is $263\text{m}$.

Conclusion:

The distance covered by the particle in $20.0\text{s}$ is $263\text{m}$.