# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 156: 49

(a) The top speed is 16.9 m/s (b) Sam travels a total distance of 230 meters.

#### Work Step by Step

(a) We can find the acceleration during the first 10 seconds. $\sum F = ma$ $F_{thrust} - F_f = ma$ $F_{thrust} - mg~\mu_k = ma$ $a = \frac{F_{thrust} - mg~\mu_k}{m}$ $a = \frac{200~N - (75~kg)(9.80~m/s^2)(0.10)}{75~kg}$ $a = 1.687~m/s^2$ We can find the speed at the end of the acceleration period which ends at $t = 10~s$. $v = a~t$ $v = (1.687~m/s^2)(10~s)$ $v = 16.9~m/s$ The top speed is 16.9 m/s (b) We can find the distance $x_1$ during the 10-second acceleration period. $x_1 = \frac{1}{2}at^2$ $x_1 = \frac{1}{2}(1.687~m/s^2)(10~s)^2$ $x_1 = 84.35~m$ We can find the rate of deceleration caused by friction after the thrust is removed. $F_f = ma$ $mg~\mu_k = ma$ $a = g~\mu_k$ $a = (9.80~m/s^2)(0.10)$ $a = 0.98~m/s^2$ We can find the time it takes for friction to bring Sam to a stop. $t = \frac{v_f-v_0}{a}$ $t = \frac{0-16.9~m/s}{-0.98~m/s^2}$ $t = 17.2~s$ We can find the distance $x_2$ while friction brings Sam to a stop. $x_2 = v_0~t+\frac{1}{2}at^2$ $x_2 = (16.9~m/s)(17.2~s)-\frac{1}{2}(0.98~m/s^2)(17.2~m/s)^2$ $x_2 = 145.7~m$ We can find the total distance $x$ that Sam travels. $x = x_1+x_2$ $x = 84.35~m+145.7~m$ $x = 230~m$ Sam travels a total distance of 230 meters.

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