## Engineering Mechanics: Statics & Dynamics (14th Edition)

a) $24~m/s$ b) $25~m/s^2$ c) $4~m/s^2$ d) $4~m/s$ e) $5~m/s^2$ f) $3~m/s$ g) $26~m$ h) $2.67~m/s$ i) $12~m/s$ j) $0.7~m/s$ and $2.1~m/s$
a) We are given $s=2t^3$ We find the velocity at $t=2~s$: $v=\frac{ds}{dt}=6t^2=6(2)^2=6*4=24~m/s$ b) We are given $v=5s$ We find the acceleration at $s=1~m$: $a=v\frac{dv}{ds}=5s(5)=25s=25*1=25~m/s^2$ c) We are given $v=4t+5$ We find the acceleration when $t=2~s$: $a=\frac{dv}{dt}=4~m/s^2$ d) We are given $a=2$ We find the velocity when $t=2$, given $v=0$ at $t=0$: $v_f=v_i+at$ $v_f=0+2(2)=0+4=4~m/s$ e) We are given $a=2$ We find $v$ when $s=4$ if $v=3$ at $s=0$ $v_f^2=v_i^2+2a(s_f-s_i)$ $v_f^2=3^2+2(2)(4-0)=9+2(2)(4)=9+4*4=9+16=25$ $v_f=5~m/s^2$ f) We are given $a=s$ We find $v$ when $s=5$ if $v=0$ at $s=4$ We know that $a=v\frac{dv}{ds}$ Thus $a~ds=v~dv$ and since $a=s$ $s~ds=v~dv$ We integrate: $\int_4^5 s~ds=\int_0^v~vdv$ $\dfrac{s^2}{2}|_4^5=\dfrac{v^2}{2}_0^v$ $v^2=5^2-4^2=25-16=9$ $v=3~m/s$ g) We are given $a=4$ We find $s$ when $t=3$ if $v=2$ and $s=2$ when $t=0$ $s=s_i+v_it+\frac{1}{2}at^2$ $s=2+2(3)+\frac{1}{2}4(3)^2=2+6+0.5(4)(9)=8+18=26~m$ h) We are given $a=8t^2$ We find $v$ when $t=1$ if $v=0$ at $t=0$ We know that $a=\frac{dv}{dt}$ Thus $\int_0^v dv=\int_0^1 a dt$ Plug in $a=8t^2$ $\int_0^v dv=\int_0^1 8t^2 dt$ $v=8(\frac{t^3}{3})_0^1=\frac{8}{3}=2.67~m/s$ i) We are given $s=3t^2+2$ We find $v$ when $t=2$ $v=\frac{ds}{dt}=6t=6(2)=12~m/s$ j) We know that the average velocity is given by: $v_a=\frac{\Delta s}{\Delta t}=\frac{7}{10}=0.7~m/s$ Note that we used the displacement, rather than the distance. To find the average speed, we use the total distance covered: $s_a=\frac{s_{total}}{\Delta t}=\frac{21}{10}=2.1~m/s$