Answer
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 $2.1~m/s$
Work Step by Step
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:
$s_a=\frac{s_{total}}{\Delta t}=\frac{21}{10}=2.1~m/s$
