Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) We will have to push for 2.0 seconds to reach the speed $v$. (b) We will have to push for 1.41 seconds for the puck with mass $2m$ to travel a distance $d$.
(a) We know that $F = ma$ Let $a_1$ be the acceleration of the puck with mass $m$. We can find an expression for $a_1$. $F = m~a_1$ $a_1 = \frac{F}{m}$ Let $a_2$ be the acceleration of the puck with mass $2m$. We can find an expression for $a_2$. $F = (2m)(a_2)$ $a_2 = \frac{F}{2m} = \frac{a_1}{2}$ We can find an expression for $v$ in terms of $a_1$ $v = a_1~t = (1.0~s)~a_1$ We can find the time for the puck with mass $2m$ to reach the speed $v$. $v = a_2~t = (1.0~s)~a_1$ $(\frac{a_1}{2})~t = (1.0~s)~a_1$ $t = 2\times (1.0~s)$ $t = 2.0~s$ We will have to push for 2.0 seconds to reach the speed $v$. (b) We can find an expression for $d$ in terms of $a_1$. $d =\frac{1}{2}~a_1t^2$ $d =\frac{1}{2}~a_1~(1.0~s)^2$ We can find the time for the puck with mass $2m$ to travel a distance $d$. $d = \frac{1}{2}a_2t^2 = \frac{1}{2}~a_1~(1.0~s)^2$ $\frac{a_1}{2}~t^2 = a_1~(1.0~s)^2$ $t = (1.0~s)~\sqrt{2}$ $t = 1.41~s$ We will have to push for 1.41 seconds for the puck with mass $2m$ to travel a distance $d$.