Chapter 6, Problem 63SP

A 1300-kg car is to accelerate from rest to a speed of 30.0 m/s in a time of 12.0 s as it climbs a 15.0° hill. Assuming uniform acceleration, what minimum power is needed to accelerate the car in this way?

To determine

The minimum power required to accelerate a car with a mass of $1300\text{ kg}$ from rest to $30\text{ m/s}$ in a duration of $12\text{ s}$ as it climbs a $15.0°$ hill.

Answer to Problem 63SP

Solution:

$98.3\text{ kW}$ or $132\text{ hp}$

Explanation of Solution

Given data:

The mass of the car is $1300\text{ kg}$.

The angle of incline of the hill is $15°$.

The time is $12\text{ s}$.

The final velocity is $30\text{ m/s}$.

Formula used:

The expression for change in kinetic energy is written as

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity and $KE$ is the change in kinetic energy.

The expression for change in potential energy is written as

$P{E}_G=mg\left(h_f-h_i\right)$

Here, $g$ is the acceleration due to gravity, $h_f$ is the final position, $h_i$ is the initial position, and $P{E}_G$ is the change in potential energy.

The expression for the work energy theorem is written as

$\text{Work done}=\text{Change in kinetic energy}+\text{change in potential energy}$

The equation of motion for the final velocity is written as

$v_f=v_i+at$

Here, $a$ is the acceleration and $t$ is the time.

The equation of motion for the displacement is written as

$s=v_{i}t+\frac{1}{2}a{t}^2$

Here, $s$ is the displacement.

The expression for average power is written as

$P=\frac{W}{t}$

Here, $P$ is the power.

Write the expression used for converting power from $\text{kW}$ to $\text{hp}$:

$P\left(\text{hp}\right)=P\left(\text{kW}\right)\left(\frac{1.34\text{ hp}}{1\text{ kW}}\right)$

Here, $P\left(\text{hp}\right)$ is power in $\text{hp}$ and $P\left(\text{kW}\right)$ is power in $\text{kW}$.

Explanation:

Draw the schematic diagram for the motion of the car:

In the above diagram, $a$ is the acceleration of the car.

The car accelerates from rest. Therefore, the initial velocity is zero.

Recall the equation of motion for the final velocity of the car:

$v_f=v_i+at$

Substitute $30\text{ m/s}$ for $v_f$, $0\text{ m/s}$ for $v_i$, and $12.0\text{ s}$ for $t$

$\begin{array}{c}30\text{ m/s}=\left(0\text{ m/s}\right)+a\left(12.0\text{ s}\right)\\ a=\frac{30\text{ m/s}}{12.0\text{ s}}\\ =2.5{\text{ m/s}}^2\end{array}$

Recall the equation of motion for the displacement of the car:

$s=v_{i}t+\frac{1}{2}a{t}^2$

Substitute $0\text{ m/s}$ for $v_i$, $12.0\text{ s}$ for $t$, and $2.5{\text{ m/s}}^2$ for $a$

$\begin{array}{c}s=\left(0\text{ m/s}\right)\left(12.0\text{ s}\right)+\frac{1}{2}\left(2.5{\text{ m/s}}^2\right){\left(12.0\text{ s}\right)}^2\\ =180\text{ m}\end{array}$

Consider the diagram. The total distance covered by the car is the sine component of the incline. Therefore,

$h_f-h_i=s\sin 15°$

Substitute $180\text{ m}$ for $s$

$\begin{array}{c}{h}_f-h_i=\left(180\text{ m}\right)\sin 15°\\ =46.59\text{ m}\end{array}$

The expression for the work energy theorem is written as

$\begin{array}{c}\text{Change in kinetic energy}+\text{change in potential energy}=\text{Work done}\\ KE+P{E}_G=W\end{array}$

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $KE$ and $mg\left(h_f-h_i\right)$ for $P{E}_G$

$W=\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)$

Substitute $46.59\text{ m}$ for $h_f-h_i$, $30\text{ m/s}$ for $v_f$, $0\text{ m/s}$ for $v_i$, $1300\text{ kg}$ for $m$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}W=\frac{1}{2}\left(1300\text{ kg}\right)\left({\left(30\text{ m/s}\right)}^2-{\left(0\text{ m/s}\right)}^2\right)+\left(1300\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\left(46.59\text{ m}\right)\\ =585,000+594,162.27\\ =1,179,162.27\text{ J}\end{array}$

Recall the expression for average power:

$P=\frac{W}{t}$

Substitute $1179162.27\text{ J}$ for $W$ and $12\text{ s}$ for $t$

$\begin{array}{c}P=\frac{1179162.27\text{ J}}{12\text{ s}}\\ =98263.52\text{ W}\left(\frac{0.001\text{ kW}}{1\text{ W}}\right)\\ =98.263\text{ kW}\\ \simeq \text{98}\text{.3 kW}\end{array}$

Recall the expression used for converting power from $\text{kW}$ to $\text{hp}$:

$P\left(\text{hp}\right)=P\left(\text{kW}\right)\left(\frac{1.34\text{ hp}}{1\text{ kW}}\right)$

Substitute $98.3\text{ kW}$ for $P\left(\text{kW}\right)$

$\begin{array}{c}P\left(\text{hp}\right)=\left(98.3\text{ kW}\right)\left(\frac{1.34\text{ hp}}{1\text{ kW}}\right)\\ =131.72\text{ hp}\\ \simeq 132\text{ hp}\end{array}$

Conclusion:

The minimum power required to accelerate the car is $98.3\text{ kW}$ or $132\text{ hp}$.