Answer :
Answer:
[tex] v_i = \sqrt{2*24.2 ft/s^2 *192 ft}= 96.40 ft/s[/tex]
And we can convert this speed to mi/h and we got:
[tex] v_i = 96.4 ft/s * 0.305 m/ft * 0.001 km/m * 0.621 mi/km *3600 s/h = 65.73 mi/h[/tex]Explanation:
For this case we know the following info given:
[tex] d = 192 ft[/tex] represent the distance
[tex]v_f =0 ft/s[/tex] represent the final velocity
[tex] \mu = 0.75[/tex] represent the kinetic friction coefficient
[tex] v_{max}= 45 mi/h[/tex] the maximum velocity
For this case from the second Law of newton we have that:
[tex] F_{net} = ma[/tex]
Assuming that we have just the friction acting we have:
[tex] - F_f = ma[/tex]
[tex] -\mu N = -\mu mg= ma[/tex]
[tex] a= -\mu g[/tex]
And replacing we have:
[tex] a = -32.2 ft/s^2 *0.75=-24.15 ft/s^2[/tex]
From kinematics we know that:
[tex] v^2_f = v^2_i +2ad[/tex]
And solving for the initial velocity we got:
[tex] v_i = \sqrt{2*24.2 ft/s^2 *192 ft}= 96.40 ft/s[/tex]
And we can convert this speed to mi/h and we got:
[tex] v_i = 96.4 ft/s * 0.305 m/ft * 0.001 km/m * 0.621 mi/km *3600 s/h = 65.73 mi/h[/tex]