Question

A world’s land speed record was set by Colonel John P. Stapp when in March 1954, he rode a rocket propelled sled that moved along a track at${}^{\mathbf{1020}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$. He and the sled were brought to a stop in${}^{\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{sec}}$. In terms of g, what acceleration did he experience while stopping?

Short answer

Acceleration in terms of ${}^{\mathrm{g}}$is${}^{21\mathrm{g}}$

Step-by-step solution

Given information

$$\begin{gathered} \mathrm{v}=1020\mathrm{km}/\mathrm{h} \\ \mathrm{t}=1.4\mathrm{s} \end{gathered}$$

Concept and formula used in the given question

The problem deals with the simple physical quantity acceleration which refers to the rate at which the velocity changes. Using the velocity and given time in standard notation, the acceleration can be found. And by dividing it bythe value of “g,” the acceleration in terms of “g” can be determined. To initiate, all the quantities should be converted to the standard units.

Formula:

The acceleration in general is given by,

$\mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}}$

To calculate acceleration in terms of g

To find acceleration, first convert${}^{1020\mathrm{km}/\mathrm{h}}$ into${}^{\mathrm{m}/\mathrm{s}}$

$$\begin{gathered} 1=\frac{\mathrm{km}}{\mathrm{h}}=\frac{1000}{3600}\mathrm{m}/\mathrm{s} \\ \frac{1020\mathrm{km}}{\mathrm{h}}\times \frac{1000}{3600}\frac{\mathrm{m}}{\mathrm{s}}\times \frac{\mathrm{h}}{1\mathrm{km}}=283.33\mathrm{m}/\mathrm{s} \\ \mathrm{v}=283.33\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (i),

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{v}}{\mathrm{t}} \\ -\frac{283.33}{1.4} \\ =202.4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Acceleration in terms of g,

$$\begin{gathered} \mathrm{a}=\frac{202.4}{9.81} \\ =21\mathrm{gm}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration in terms of${}^{\mathrm{g}}$is${}^{21\mathrm{g}}$