Question

A hot rod can accelerate from $\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{60}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}\mathbf{}\mathbf{in}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{s}\mathbf{.}$(a) What is its average acceleration, in $\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, during this time? (b) How far will it travel during the, assuming its acceleration is constant? (c) From rest, how much time would it require to go a distance of 0.25 kmif its acceleration could be maintained at the value in (a)?

Short answer

1. The average acceleration of the rod is $3.1\mathrm{m}/{\mathrm{s}}^2$
2. Distance traveled by the rod in $5.4\mathrm{is}45\mathrm{m}.$
3. Time required to travel$0.25\mathrm{km}\mathrm{is}13\sec .$

Step-by-step solution

Given data

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{quantities}\mathrm{are}\mathrm{as}\mathrm{follows}. \\ \mathrm{I}\mathrm{nitial}\mathrm{Velocity}\left({\mathrm{v}}_0\right)=0\mathrm{km}/\mathrm{hr}=0\mathrm{m}/\mathrm{s} \\ \mathrm{Final}\mathrm{Velocity}\left(\mathrm{v}\right)=60\mathrm{km}/\mathrm{hr} \\ \mathrm{I}\mathrm{nitial}\mathrm{Time}\left({\mathrm{t}}_0\right)=0\sec \\ \mathrm{Final}\mathrm{Time}\left({\mathrm{t}}_{\mathrm{final}}\right)=5.4\sec \end{gathered}$$

Understanding the kinematic equations of motion

Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion. Using the kinematic equations, you can relate the displacement, velocity, acceleration, and time.

(a) Calculations for average accelerations

$3.1\mathrm{m}/{\mathrm{s}}^2$Convert the velocity into m/s.

$$\begin{gathered} \mathrm{Final}\mathrm{Velocity}\left(\mathrm{v}\right)=60\frac{\mathrm{km}}{\mathrm{hr}} \\ =60\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1\mathrm{hr}}{3600\sec }\times \frac{1000\mathrm{m}}{1\mathrm{km}} \\ =16.67\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, calculate the acceleration using the formula for acceleration.

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{Final}\mathrm{velocity}-\mathrm{Initial}\mathrm{velocity}}{\mathrm{Initial}\mathrm{Time}-\mathrm{Final}\mathrm{Time}} \\ =\frac{16.67-0}{54-0} \\ =3.08 \\ \approx 3.1\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value for average acceleration is$3.1\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculations for distance travelled by hot rod

The kinematic equation for the displacement is,

$\mathrm{x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (i)

Substitute the given values in the equation (i) to calculate displacement.

$$\begin{gathered} \mathrm{x}=0+\frac{1}{2}\times 3.1\times 5.4^2 \\ =\frac{90.396}{2} \\ =45.19 \\ \approx 45\mathrm{m} \end{gathered}$$

Hence, the distance travelled by the hot rod can be given as 45 m

(c) Calculation for time

The distance covered is, $\mathrm{x}=0.25\mathrm{km}$Converting the distance into m

role="math" localid="1656304283895" $$\begin{gathered} \mathrm{x}=0.25\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}} \\ =250\mathrm{m} \\ \mathrm{a}=3.1\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{I}\mathrm{nitial}\mathrm{velocity}\left({\mathrm{v}}_0\right)=0\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Substitute the values in equation (i) again to calculate time.

$$\begin{gathered} 250=0+\frac{1}{2}\times 3.1\times {\mathrm{t}}^2 \\ 250=1.55{\mathrm{t}}^2 \\ {\mathrm{t}}_2=\frac{250}{1.55} \\ \mathrm{t}=12.7 \\ \approx 13\sec \end{gathered}$$

So, the time required to go a distance of 0.25 km if its acceleration could be maintained at the value in (a) is 13 s.