Q54E - University Physics with Modern Physics

Question

High-speed motion pictures ( 3500 frames/second)2 of a jumping, $\mathbf{210}\mathbf{-}\mathbf{\mu g}$flea yielded the data used to plot the graph in Fig. E2.54. (See “The Flying Leap of the Flea” by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about  long and jumped at a nearly vertical takeoff angle. Use the graph to answer these questions: (a) Is the acceleration of the flea ever zero? If so, when? Justify your answer. (b) Find the maximum height the flea reached in the first  2.5ms. (c) Find the flea’s acceleration at 0.5 ms, 1.0 ms , and 1.5 ms. (d) Find the flea’s height at 0.5 ms,  1.0 ms , and 1.5 s.

Step-by-Step Solution

Answer

a) the acceleration of the flea will be zero at $\mathrm{t}=1.25\hspace{0.33em}\mathrm{ms} \mathrm{to}\hspace{0.33em}\mathrm{t}=2.5\hspace{0.33em}\mathrm{ms}$, b) the maximum height which is reached by the flee is , c) the acceleration of the flea at $0.5 ms,1.0 ms and 1.5 ms$ is  $5\times {10}^4 cm/s$, $9.38\times {10}^4 cm/s$ and 0 respectively and d) the height of the flee at $0.5 ms,1.0 ms and 1.5 ms$ are $0.00012\hspace{0.33em}cm$,  $0.0005\hspace{0.33em}\mathrm{cm}$ and 0.0011 cm respectively.

Step 1: Identification of the given data

The given data can be expressed below as:

• The speed of the high-speed motion pictures is ( 3500 frames/second)  .
• The data has been yielded by $210-\mu g$ flea.
• The flea is about 2mm long.

Step 2: Significance of Newton’s first law for the flea

This law states that a body will continue to be in uniform motion or velocity unless it is acted by an external force.

The equation of motion will be helpful to find out a graph to deduce the acceleration and the maximum height. Moreover, the flea’s acceleration and height will be deduced with the help of the equation.

Step 3: Determination of the acceleration, maximum height, acceleration at different points and height of the flea

a) the rate of change of acceleration can be expressed as:

$a=\frac{dv}{dt}$

Here is the change of the velocity and is the change of time.

The graph may be helpful for making the acceleration of the flea zero which has been produced below-

Here, in the $v_y-t$ graph, the line BC is constant during $t=1.25\hspace{0.33em}ms to t=2.5\hspace{0.33em}ms$, hence, the acceleration will be zero during the particular period.

Thus, the acceleration of the flea will be zero at $t=1.25\hspace{0.33em}ms to\hspace{0.33em}t=2.5\hspace{0.33em}ms$.

b) the formula for calculating the maximum height is as follows-

$$\begin{gathered} v_{xf}^2-v_{yi}^2=2gh \\ h=\frac{v_{xf}^2-v_{yi}^2}{2g} \end{gathered}$$.. i)

Here, $v_{yi} and\hspace{0.33em}{v}_{xf}$ are the initial and the final velocity of the flee respectively and h is the maximum height reached by the flee.

Analysing the data from the graph, the final velocity of the flee is zero and the initial velocity of the flee is 130 cm/s.

Substituting the values in the equation i), we get-

$$\begin{gathered} h=\frac{v_{xf}^2-v_{yi}^2}{2g} \\ h=\frac{(0\mathrm{m}/\mathrm{s}{)}^2-\left(130\mathrm{cm}/\mathrm{s}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)}{2\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)} \\ h=8.62\mathrm{cm} \end{gathered}$$

Thus, the maximum height which is reached by the flee is 8.62 cm.

c) the acceleration of the flee at 0.5 ms is expressed as:

$a_{0.5ms}=\frac{d{v}_1}{d{t}_1}$

Here, ${\mathrm{dv}}_1\hspace{0.33em}\mathrm{and}\hspace{0.33em}{\mathrm{dt}}_1$ are the velocity and the time of the flee at 0.5 ms.

Hence, taking the value of $d{v}_{1}25\hspace{0.33em}cm/s$ from the graph and the value of $d{t}_{1}0.5\hspace{0.33em}ms$, we get-

$$\begin{gathered} a_{0.5\mathrm{ms}}=\frac{25\mathrm{cm}/\mathrm{s}}{0.5\times {10}^{-3}\mathrm{s}} \\ =5\times {10}^4\mathrm{cm}/\mathrm{s} \end{gathered}$$

the acceleration of the flee at 1.0 ms is expressed as:

$a_{1.0ms}=\frac{d{v}_2}{d{t}_2}$

Here, $d{v}_2\hspace{0.33em}and d{t}_2$ are the velocity and the time of the flee at 1.0 ms.

Hence, taking the value of from the graph and the value of $d{t}_2$ is 0.5 ms, we get-

$$\begin{gathered} a_{1.0\mathrm{ms}}=\frac{93.8\mathrm{cm}/\mathrm{s}}{1.0\times {10}^{-3}\mathrm{s}} \\ =9.38\times {10}^4\mathrm{cm}/\mathrm{s} \end{gathered}$$

Analysing the above figure, the acceleration of the flea at 1.5 ms is zero.

As the velocity at 1.5 ms is zero, so the acceleration also becomes zero at that point.

Thus, the acceleration of the flea at $0.5 ms,1.0 ms an 1.5 ms$ is $5\times {10}^4 \mathrm{cm}/\mathrm{s} ,9.38\times {10}^4 \mathrm{cm}/\mathrm{s}$, and 0 respectively. 

d) the height of the flea at 0.5 ms is expressed as:

$h_{0.5 ms}=\frac{1}{2}g{t}^2$

The value of g is $980\hspace{0.33em}cm/s^2$, t is the time taken of the flea which is 0.5 ms. 

Substituting the values in the above expression, we get-

$$\begin{gathered} h_{0.5\mathrm{ms}}=\frac{1}{2}\left(980\mathrm{cm}/{\mathrm{s}}^2\right){\left(0.5\times {10}^{-3}\mathrm{s}\right)}^2 \\ =0.00012\mathrm{cm} \end{gathered}$$

the height of the flea at 1.0 ms is expressed as:

$h_{1.0 ms}=\frac{1}{2}g{t}^2$

The value of g is $980\hspace{0.33em}cm/s^2$, t is the time taken of the flea which is .1.0 ms 

Substituting the values in the above expression, we get-

$$\begin{gathered} h_{1.0\mathrm{ms}}=\frac{1}{2}\left(980\mathrm{cm}/{\mathrm{s}}^2\right){\left(1.0\times {10}^{-3}\mathrm{s}\right)}^2 \\ =0.0005\mathrm{cm} \end{gathered}$$

the height of the flea at 1.5 ms is expressed as:

$h_{1.5 ms}=\frac{1}{2}g{t}^2$

The value of g is $980\hspace{0.33em}cm{s}^2$ , t is the time taken of the flea which is .1.5 ms 

Substituting the values in the above expression, we get-

$$\begin{gathered} h_{1.5\mathrm{ms}}=\frac{1}{2}\left(980\mathrm{cm}/{\mathrm{s}}^2\right){\left(1.5\times {10}^{-3}\mathrm{s}\right)}^2 \\ =0.0011\mathrm{cm} \end{gathered}$$

Thus, the height of the flee at $0.5 \mathrm{ms},1.0 \mathrm{ms} \mathrm{and} 1.5 \mathrm{ms}$ are 0.00012 cm, 0.0005 cm and 0.0011 cm respectively.