SOLUTION:
Draw the free-body diagram of the car
Consider the vertical direction
Consider the motion in the horizontal direction
Solve for the acceleration of the car.
Solve for the coefficient of kinetic friction
Solution:
We are given the following values:v_0=0\:\text{m/s}; v_f=65.0\:\text{m/s}; and \Delta x=0.250\:\text{m}.
We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.
Part B
Solve for the acceleration first using the formula
\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x
We solve for acceleration in terms of the other variables.
a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}Substitute the given values
\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\
a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part A
To solve for the time of this motion, we shall use the formula
v_f=v_0+at
Solving for time, t, in terms of the other variables we have.
t=\frac{v_f-v_0}{a}We now substitute the values given, and the computed acceleration to find the time.
\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\
t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Solution:
We are given the following: v_0=0 \ \text{m/s}; v_f=26.8 \ \text{m/s}; and t=3.90\ \text{s}.
Part A
The average acceleration of the motorcycle can be solved using the equation \overline{a}=\frac{\Delta v}{\Delta t}. Substitute the given into the equation. That is,
\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\
\overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
The distance traveled is equal to the average velocity multiplied by the time of travel. That is,
\begin{align*}
\Delta x & =v_{ave}t\\
\Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\
\Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}Solution:
The best equation that can be used to solve this problem is \Delta x=v_{ave} t. That is,
\begin{align*}
\Delta x & = v_{ave} t \\
\Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\
\Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Therefore, the distance over which the puck accelerates is 0.7992 meters.
Solution:
Part A
The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.
Part B
The list of known variables are:
Initial velocity: v_0=0\:\text{m/s}
Final Velocity: v_f=30.0\:\text{cm/s}
Distance Traveled: x-x_0=1.80\:\text{cm}
Part C
The best equation to solve for this is \Delta \text{x}=\text{v}_{\text{ave}}\text{t} where v_{ave} is the average velocity, and t is time. That is
\begin{align*}
\Delta x & =v_{ave} t \\
t &=\frac{\Delta x}{v_{ave}} \\
t & =\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}\\
t & =0.12\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}Part D
Since the computed value of the time for the acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.
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