Question

An automobile traveling on a highway has an average kinetic energy of . Its mass is . What is its average speed? Convert your answer to miles per hour to see whether it makes sense. If you could use all of themc²rest energy of some amount of fuel to provide the car with its kinetic energy of , What mass of fuel would you need?

Short answer

The average speed of the automobile is $12M{S}^{-1}$.

The speed of the automobile in miles per hour is 26.85 Miles/h .

The mass of fuel needed is $1.22\times {10}^{12}kg$.

Step-by-step solution

Definition of kinetic energy

The kinetic energy is given by:

$k=\frac{1}{2}m{v}^2$

Here mis the mass of the object, and vis the velocity of the object.

A simple approximate expression for the kinetic energy of the car can be obtained by comparing the speed v of a particle to the speed of light $v<<c$.

Determine the velocity of the automobile

Let vbe the automobile's speed.

In the given scenario, the kinetic energy is negligible compared to the rest energy, with the kinetic energy being approximately zero at low speeds.

Solve the formula of kinetic energy for v.

$v=\sqrt{\frac{2K}{m}}$

Substitute the kinetic energy of the automobile$k=1.1\times {10}^5$and its mass$m=1500kg$ into the obtained formula,

$$\begin{gathered} v=\sqrt{\frac{2\times \left(1.1\times {10}^{15}J\right)}{1500kg}} \\ =\sqrt{\frac{2\times \left(1.1\times {10}^5\right)\left(1J\right)\times \left({\displaystyle \frac{1kg{\left(m/s\right)}^2}{1J}}\right)}{1500\left(1kg\right)}} \\ =12.0m/s \end{gathered}$$

Convert the units of the speed into mile/h by using the conversion $1mile=1609m$ and $1h=3600s$.

$$\begin{gathered} v=12.0\left(\frac{m}{s}\times \left[\frac{1mile}{1609m}\right]\times \left[\frac{3600s}{1h}\right]\right) \\ 26.85miles/h \end{gathered}$$

Thus, the average speed is$12.0m/s$ and in miles per hour the speed is $26.85miles/h$.

Calculation for the mass of fuel

The rest energy of the fuel is given by:

.$E_{rest}=m{c}^2$

Substitute $E_{rest}=1.1\times {10}^{5}J$and $c=3\times {10}^{8}m/s$into the formula and solve for m.

role="math" localid="1654159452050" $$\begin{gathered} 1.1\times {10}^{5}J=m\left(3\times {10}^{8}m/s\right) \\ m=\frac{1.1\times {10}^5\left(1J\right)\times \left({\displaystyle \frac{1kg{\left(m/s\right)}^2}{1J}}\right)}{{\left(3\times {10}^8\right)}^2{\left(1m/s\right)}^2} \\ =1.22\times {10}^{-12}kg \end{gathered}$$

Thus, the mass of fuel needed is$1.22\times {10}^{-12}kg$.

As illustrated, a small amount of fuel is sufficient to propel the vehicle at a speed of $26.85miles/h$.