Q45E - University Physics with Modern Physics

Question

The rocket-driven sled Sonic Wind No. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s (500 mi/h) in 0.900 s. 

(a) Compute the acceleration in $\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, assuming that it is constant. 

(b) What is the ratio of this acceleration to that of a freely falling body (g)? (c) What distance is covered in 0.900 s? 

(d) A magazine article states that at the end of a certain run, the speed of the sled decreased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40g. Are these figures consistent?

Step-by-Step Solution

Answer

(a) The acceleration of the rocket is $248.9\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$

(b) The ratio between rocket acceleration and acceleration due to gravity is 25.4.

(c) The distance covered in 0.900 s is 100.80 m.

(d) The figures are inconsistent.

Step 1: Given data

• The initial velocity of the rocket; $u=0\hspace{0.33em}m/s$.
• The final velocity of the rocket; $v=224\hspace{0.33em}m/s$.
• Time taken by the rocket to achieve the final speed; $t=0.900\hspace{0.33em}s$.

Step 2: (a) Determination of the acceleration

Newton’s first law of motion can be expressed as,

 

$v=u+at$                                                                                  ……………………… (1)

 

Here v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

 

Substituting the given data in the above expression, we get

$$\begin{gathered} 224\mathrm{m}/\mathrm{s}=0\mathrm{m}/\mathrm{s}+a\times 0.900\mathrm{s} \\ \mathrm{a}=\frac{224\mathrm{m}/\mathrm{s}}{0.900\mathrm{s}} \\ \mathrm{a}=248.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

Thus, the acceleration of the rocket is $248.9\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$

Step 3: (b) Determine the ratio of this acceleration

The acceleration of a freely falling body is $g=9.8\hspace{0.33em}m/s^2$.

 

Therefore, the ratio between rocket acceleration and acceleration due to gravity is, 

 

$$\begin{gathered} \frac{a}{g}=\frac{248.9\mathrm{m}/{\mathrm{s}}^2}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =25.39 \end{gathered}$$

 

Thus, the ratio between rocket acceleration and acceleration due to gravity is 25.39.

Step 4: (c) Determine the distance is covered in 0.900s

Newton’s first law of motion can be expressed as,

 

$\mathbf{d}\mathbf{=}\mathbf{ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$                                                                       ………………………… (2)

d is the distance traveled.

 

Substituting the given data in the above expression, we get:

 

$$\begin{gathered} d=(0\mathrm{m}/\mathrm{s})\times \left(0.900\mathrm{s}\right)+\frac{1}{2}\left(248.9\mathrm{m}/{\mathrm{s}}^2\right)(0.900\mathrm{s}{)}^2 \\ =100.80\mathrm{m} \end{gathered}$$

 

Thus, the distance covered in 0.900 s is 100.80 m.

Step 5: (d) Determination of the magazine figure’s consistency

• The initial speed of the sled is 283 m/s
• The final speed of the sled is 0 m/s 
• The time duration is 1.40 s

 

Substituting these values in the equation (1), we get

 

$$\begin{gathered} a=\frac{-283\mathrm{m}/\mathrm{s}}{1.40\mathrm{s}} \\ a=-202.1\mathrm{m}/{\mathrm{s}}^2 \\ \frac{\mathrm{a}}{\mathrm{g}}=\frac{-202.1\mathrm{m}/{\mathrm{s}}^2}{-9.8\mathrm{m}/{\mathrm{s}}^2} \\ \frac{\mathrm{a}}{\mathrm{g}}=20.6 \end{gathered}$$

 

This acceleration value is almost half of their given value of 40 g.

 

Thus, the figures are inconsistent.