Calculating the unit and value of k given a velocity function| University Physics

PART A. Determine the units of k in terms of m and s.

ANSWER: $m/s^3$

We know that the unit for velocity is m/s and the unit for time is s. Therefore,

$v=kt^2$

$m/s=k\left(s\right)^2$

$k=\frac{m/s}{s^2}$

$k=m/s^3$

PART B. Determine the value of the constant k.

ANSWER: $k=49.8\:m/s^3$

$v_x=\frac{dx}{dt}$

$dx=v_xdt$

$\int \:dx=\int \:v_xdt$

$x=\int \:kt^2dt$

$x=\frac{kt^3}{3}+C$

Solve for C using the pairs $x=-7.90\:m,\:\:t=0\:s$

$C=-7.9\:$

The position function therefore is

$x\left(t\right)=\frac{kt^3}{3}-7.90$

Solve for k using the pair $x=8.70\:m,\:t=1.00\:s$

$8.70=\frac{k\left(1\right)^3}{3}-7.90$

$\frac{k}{3}=8.70+7.90$

$k=3\left(16.6\right)$

$k=49.8\:m/s^3$