Answer:
The speed of the automobile after 1.25 s have elapsed is 8.65 m/s
Explanation:
Given;
initial velocity of the driver, [tex]v_i[/tex] = 16 m/s
coefficient of kinetic friction, [tex]\mu_k[/tex] = 0.6
time of motion, t = 1.25s
The final velocity of the driver is calculated as follows;
[tex]a = \frac{\Delta v}{\Delta t} = - \frac{v_f - v_i}{\Delta t}\\\\ (the \ negative \ sign \ indicates \ that \ the \ final \ velocity \ will \,be\ smaller \\\ than \ the \ initial \ velocity \ because \ of \ limiting \ frictiona\ force)\\\\v_f- v_i = -a\Delta t\\\\v_f = v_i - at\\\\Recall \ from \ Newton's \ law; \ F_k = \mu_k mg = ma \\ \ a = \mu_kg\\\\v_f = v_i - (\mu_k g )t\\\\v_f = 16 - (0.6 \times 9.8) 1.25\\\\v_f = 16 - 7.35\\\\v_f = 8.65 \ m/s[/tex]
Therefore, the speed of the automobile after 1.25 s have elapsed is 8.65 m/s
