Question

A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. what is the average power delivered by the engine? (1 hp = 746 w)

Answer 1

The solution for this is:
Work done = force * distance = m*a*d and power = energy/time 
The vo=0 and vf = 25 m/s and t=7 sec. This gives... 
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N. 
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again). 

Answer 2

The average power delivered by the engine is $67500\text{ watt}$ or $90\text{ hp}$.

Further explanation:

Power is defined as the rate at which work is transferred. Its SI unit is watt or J/s. Average power is the average amount of work transferred per unit time.  

First we calculate the average velocity of the car then after the distance traveled by car in $7\text{ s}$ and average power delivered can be determined.

Concept:

Work is defined as the dot product of force and displacement. Si unit of work is joule which is the work done by a force of 1 N in moving an object through a distance of 1 m in the direction of force.

Write the expression for average velocity as:

${V_{{\text{avg}}}} = \dfrac{{{V_{\text{i}}} + {V_{\text{f}}}}}{2}$

Here, ${V_{\text{i}}}$ is the initial velocity of the car and ${V_{\text{f}}}$ is the final velocity of the car.

Substitute $0$ for ${V_{\text{i}}}$ and $25\text{ m/s}$ for ${V_{\text{f}}}$ in the above expression.

$\begin{aligned}{V_{{\text{avg}}}}&=\frac{{0+25\,{\text{m/s}}}}{2}\\&=12.5\text{ m/s}\\\end{aligned}$

Distance travelled by car in $7\text{ s}$ is determined by the following relation:

$d = {V_{\text{avg}}\cdot t$

Substitute $12.5\text{ m/s}$ for ${V_{{\text{avg}}}}$ and $7\text{ s}$ for $t$ in the above expression.

$\begin{aligned}d&=({1.5\,{\text{m/s}}})({7\,{\text{s}}})\\&=87.5\text{ m}\\\end{aligned}$

Acceleration of the car in time interval of $7\text{ s}$,

$\begin{aligned}a&=\frac{{\Delta V}}{{\Delta t}}\\&=\frac{{{V_{\text{f}}} - {V_{\text{i}}}}}{{\Delta t}}\\\end{aligned}$

 

Substitute values for ${V_{\text{f}}}$, $0$ for ${V_{\text{i}}}$ and $7\text{ s}$ for $\Delta t$ in the above expression.

$\begin{aligned}a&=\frac{{25-0}}{7}\\&=3.6\,{\text{m/}}{{\text{s}}^{\text{2}}}\\\end{aligned}$

From newtons second law,

$F = ma$

Substitute $1500\text{ kg}$ for $m$ and $3.6\text{ m/s}^2$ for $a$ in the above expression.

$\begin{aligned}F&=\left( {1500\,{\text{kg}}} \right)\left( {3.6\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\\&=5400\,{\text{N}}\\\end{aligned}$

Write the expression for work transferred,

$\boxed{W = Fd}$

Substitute $5400\text{ N}$ for $F$ and $87.5\text{ m}$ for $d$ in the above expression.

$\begin{aligned}W&=\left( {5400\,{\text{N}}} \right)\left( {87.5\,{\text{m}}} \right)\\&=4.73 \times {10^5}\,{\text{J}}\\\end{aligned}$

Average power delivered by the engine is the work transfer per unit time.

$\boxed{P = \dfrac{W}{t}}$

Here, $P$ is the power delivered, $W$ is the work transfer and $t$ is the time interval.

Substitute $4.73 \times {10^5}\,{\text{J}}$ for $W$ and $7\text{ s}$ for $t$ in the above expression.

$\begin{aligned}P&=\frac{{\left( {4.73 \times {{10}^5}\,{\text{J}}} \right)}}{{\left( {7\,{\text{s}}} \right)}}\\&=67570\,{\text{W}}\\\end{aligned}$

Converting unit of Watt in hp,

Given that,   $\boxed{1\,{\text{hp}} = 746\,{\text{W}}}$

$\begin{aligned}P&=\left( {67570\,{\text{W}}} \right)\left( {\frac{{1\,{\text{hp}}}}{{746\,{\text{W}}}}} \right) \\&=90.5\,{\text{hp}}\\\end{aligned}$

Thus, the average power delivered by the engine is $67500\text{ watt}$ or $90\text{ hp}$.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

1500 kg, accelerate, 25 m/s, 25 meter per second, 7.0 s, average power, average velocity, distance, rate of work, force, momentum change, horse power, 90 hp, 90.5 hp, 67500 W, 67570 W, 4.7 times 10^5 j, 3.6 meter per sec^2, 3.6 m/s^2.