Respuesta :
Answer:
a) [tex] a_A = \frac{2* (-10ft)}{(2.5s)^2}= -3.2 \frac{ft}{s^2}[/tex]
[tex] 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}[/tex]
b) For this case we can use this formula:
[tex] t= \frac{V_f - V_i}{a}[/tex]
And replacinf for the runner B we got:
[tex] t= \frac{22 ft/s -0 ft/s}{3.72 ft/s^2}= 5.91 s[/tex]
And the runner B should begin to run at:
[tex] t_{B start}= 5.91-2.5 s = 3.41 sec[/tex]
Explanation:
For this case we have the following info given:
[tex] V_{iA}= 30 ft/s[/tex] represent the initial velocity for runner A
[tex] V_{iB}= 0 ft/s[/tex] represent the initial velocity for runner B
[tex] \Delta t = 2.5 s[/tex] represent the difference of time between the two runners
[tex] \Delta x= 65 ft[/tex] represent the total distance for the exchange zone.
Part a
For this case we can use the following formula:
[tex]x_f = x_i + v_i t +\frac{1}{2}at^2[/tex]
And we can define [tex] \Delta x = x_f -x_i [/tex] and we can convert this equation into:
[tex] \Delta x= v_i t +\frac{1}{2}at^2[/tex]
And we can find the acceleration since we have all the other values like this:
[tex] 65 ft = 30\frac{ft}{s} (2.5s) + \frac{1}{2}a_A (2.5s)^2[/tex]
[tex] -10 ft =\frac{1}{2}a_A (2.5s)^2[/tex]
[tex] a_A = \frac{2* (-10ft)}{(2.5s)^2}= -3.2 \frac{ft}{s^2}[/tex]
And that would be the acceeleration for the runner A.
For the acceleration of the runner A we need to take in count that [tex] V_{fA}= V_{fB}[/tex]
And for this case we can use this:
[tex]V_{fB}= V_{fA}= V_{iA}+ a_A t[/tex]
And if we replace we got:
[tex]V_{fB}= V_{fA}= 30 \frac{ft}{s}+ (-3.2 \frac{ft}{s^2}) (2.5s) = 22 \frac{ft}{s}[/tex]
So for this case we have [tex] V_{iB}= 0ft/s , V_{fB}= 22 ft/s, \Delta x= 65 ft[/tex], and we can use this formula:
[tex] V^2_f = V^2_i + 2 a \Delta x[/tex]
And solving for a we got:
[tex] 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}[/tex]
Part b
For this case we can use this formula:
[tex] t= \frac{V_f - V_i}{a}[/tex]
And replacinf for the runner B we got:
[tex] t= \frac{22 ft/s -0 ft/s}{3.72 ft/s^2}= 5.91 s[/tex]
And the runner B should begin to run at:
[tex] t_{B start}= 5.91-2.5 s = 3.41 sec[/tex]
