Question

As relay runner A enters the 65-ft-long exchange zone with a speed of 30 ft/s, he begins to slow down. He hands the baton to runner B 2.5 s later as they leave the exchange zone with the same velocity. Determine (a) the uniform acceleration of each of the runners, (b) when runner B should begin to ru

Answer 1

Answer:

a) $a_A = \frac{2* (-10ft)}{(2.5s)^2}= -3.2 \frac{ft}{s^2}$

$a_B = \frac{V^2_{fB} -V^2_{iB}}{2 \Delta x}= \frac{(22ft/s)^2 -(0)}{2*65 ft}= 3.72 \frac{ft}{s^2}$

b) For this case we can use this formula:

$t= \frac{V_f - V_i}{a}$

And replacinf for the runner B we got:

$t= \frac{22 ft/s -0 ft/s}{3.72 ft/s^2}= 5.91 s$

And the runner B should begin to run at:

$t_{B start}= 5.91-2.5 s = 3.41 sec$

Explanation:

For this case we have the following info given:

$V_{iA}= 30 ft/s$ represent the initial velocity for runner A

$V_{iB}= 0 ft/s$ represent the initial velocity for runner B

$\Delta t = 2.5 s$ represent the difference of time between the two runners

$\Delta x= 65 ft$ represent the total distance for the exchange zone.

Part a

For this case we can use the following formula:

$x_f = x_i + v_i t +\frac{1}{2}at^2$

And we can define $\Delta x = x_f -x_i$ and we can convert this equation into:

$\Delta x= v_i t +\frac{1}{2}at^2$

And we can find the acceleration since we have all the other values like this:

$65 ft = 30\frac{ft}{s} (2.5s) + \frac{1}{2}a_A (2.5s)^2$

$-10 ft =\frac{1}{2}a_A (2.5s)^2$

$a_A = \frac{2* (-10ft)}{(2.5s)^2}= -3.2 \frac{ft}{s^2}$

And that would be the acceeleration for the runner A.

For the acceleration of the runner A we need to take in count that $V_{fA}= V_{fB}$

And for this case we can use this:

$V_{fB}= V_{fA}= V_{iA}+ a_A t$

And if we replace we got:

$V_{fB}= V_{fA}= 30 \frac{ft}{s}+ (-3.2 \frac{ft}{s^2}) (2.5s) = 22 \frac{ft}{s}$

So for this case we have $V_{iB}= 0ft/s , V_{fB}= 22 ft/s, \Delta x= 65 ft$, and we can use this formula:

$V^2_f = V^2_i + 2 a \Delta x$

And solving for a we got:

$a_B = \frac{V^2_{fB} -V^2_{iB}}{2 \Delta x}= \frac{(22ft/s)^2 -(0)}{2*65 ft}= 3.72 \frac{ft}{s^2}$

Part b

For this case we can use this formula:

$t= \frac{V_f - V_i}{a}$

And replacinf for the runner B we got:

$t= \frac{22 ft/s -0 ft/s}{3.72 ft/s^2}= 5.91 s$

And the runner B should begin to run at:

$t_{B start}= 5.91-2.5 s = 3.41 sec$