Physics Bootcamp

(a) Since Karl's speed is greater, he will finish first.

(b) Using \(v = d/t \) on Karl gives

(c) Using \(v = d/t \) on Justin gives

(d) Since Karl finishes at \(t= 50\text{ min} \text{,}\) Justin will have covered the following distance by then (note: I need to convert \(50\text{ min} = 5/6\text{ hr} \) since the speed is given in per \(\text{hr}\) unit.)

That means, the remaining distance is

(e) Using \(d = v t \) with \(v\) being the difference of the two speeds. Also, since the speeds have time in \(\text{hr}\text{,}\) we need to convert \(t\) to \(\text{hr} \) as well.

You can also do this part by first finding how far Karl and Justin went, and then subtracting their distances.

(a) Our answers will differ since we may break up the area between \(t=0\) and \(t=30\text{ sec}\) in different ways. My area will come from the areas shaded in Figure 2.9.29.

The areas sum up to

(b) Our answers will differ since we may draw tangents a little differently. Figure 2.9.30 contains the tangents at the four instants.

The slopes of these tangents give the acceleration at the corresponding instants.

The value at \(t= 30\text{ sec}\) is negative because the cheetah, while moving towards positive \(x \) axis, is slowing down.

(a) Since the motion does not have a turn around in direciton, the speed plot is same as velocity plot. Therefore, we can get the distance traveled by area-under-the-curve. We also convert the time into seconds.

(b) From the slopes we find three values of accelerations.

We plot these in Figure 2.9.33.

(c) (i) With \(a_{x} = \dfrac{1}{3}\text{ m/s}^2\text{,}\) \(v_{i,x}=0\text{,}\) and \(\Delta t = 120\text{ sec}\text{,}\) we get

(ii) This is a constant velocity motion.

(iii) Finally, in this part, we have \(a_{x} = -\dfrac{2}{3}\text{ m/s}^2\text{,}\) \(v_{i,x}=40\text{ m/s}\text{,}\) and \(\Delta t = 60\text{ sec}\text{.}\) These give us

Therefore, the total

(a), (b) , (c), all three parts can be solved by setting up two sets of equations, one set for each stone, and relation \(t_1 = t_2 + 0.5 \) for the times of the two stones. Let us use positive \(y\) axis up to keep track of signs with origin at the place the motions started. That will make \(y_i = 0\) and \(y_f = -50 \text{ m}\text{.}\) Let us use subscript 1 for stone 1 and 2 for stone 2, and drop \(y\) from subscripts.

For stone 1, we have the following two equations.

For stone 2, we have the following two equations.

with

From Eq. (2.9.1), we solve for \(t_1\) and keep the positive root as our answer.

Therefore,

Using \(t_1\) in Eq. (2.9.2) give,

Using \(t_2\) in Eq. (2.9.3) gives

Using \(t_2\) and \(v_{2i}\) in Eq. (2.9.4) gives

Let \(y=0\) at the launch point with positive \(y\) axis pointed up. The first ball reaches \(y=h\) before falling with

Now, ball 1 moves in the negative \(y\) direction at \(t=0\) starting at \(y_{1i}=20.39\text{ m}\) with \(v_{1i}=0\) and ball 2 moves in the positive \(y\) direction \(y_{2i}=0\) starting with \(v_{2i}=20\text{ m/s}\text{.}\) We want \(y\) where they hit each other, i.e., \(y_{1f} = y_{2f} = y\text{.}\) Let \(t\) be the duration since ball 2 was launched. For the two equations we get

Subtracting first equation from the second we get

Using this \(t\) we get

(a) and (b):

There are two constant-acceleration segments in this problem, such that final velocity and final position of the first segment is the initial velocity and initial position of the second segment. Let us denote the quantities of the first and second segments by subscripts 1 and 2, respectively.

We point positive \(x \) axis from the initial point of the first segment towards the final point where the puck comes to rest. Using this coordinate system, we have the following knowns and unknowns for segment 1.

First Segment:

Suppressing units in calculations, we list knowns and unknowns.

The equation \(v_f^2 = v_i^2 + 2 a x \) gives

This gives

Since the puck is moving in the same direction as the direction of the positive \(x \) axis, the positive root is the right choice here.

Now, using \(v_f = v_i + a t\text{,}\) we get

Second Segment:

Let us use \(t_2 \) for \(t_{2,f} - t_{2,i}\) and place another origin at \(x_{2,i}\text{.}\)

Using \(v_f = v_i + a t\text{,}\) we get \(t\)

Using \(x = v_i t +\dfrac{1}{2}at^2\text{,}\) we get

(a) The total distance is \(50 + 54.3 = 104.3\text{ m}\text{.}\)

(b) The total time is

(a) We would find velocity by integrating the acceleration, and then get the speed from the absolute value. The magnitude of acceleration drops from \(g \) to zero in time \(\tau\text{.}\) Therefore, we have the following formula for the magnitude of the acceleration at instant \(t\text{.}\)

Let positive \(y \) axis be pointed up so that

Now, we integrate this from \(t = 0\) to \(t=t\text{.}\) With \(v_y=0\) at \(t = 0\text{.}\)

Therefore, speed at arbitrary instant is

(b) Speed at instant \(\tau\) will be obtained by setting \(t= \tau\) in the answer for part (a) and taking the absolute value.

(c) Integrating \(v_y(t) \) will give \(\Delta y\text{.}\)

The magnitude of this gives the distance fallen.

(a) Setting \(a_y=0\) and \(v_y = v_T \text{,}\) we imediately get the answer.

(b) We express \(a_y = dv_y/dt\) and rearrange to obtain

Integrating this from \(t=0\) to \(t =t\) with \(v_y(0)=0\) and \(v_y(t)=v_y\text{,}\) we get

Since \(g\) is larger than \(\gamma v_y \text{,}\) exponentiating and then rearranging gives.

which is

(c) Integrating \(v_y\) gives the desired answer.

Use \(v_y \) from (b) and carry out the integration.

No solution provided.

No solution provided.

No solution provided.