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

A particle’s velocity is described by the function v_x=kt^2 where v_x is in m/s, t is in s, and k is a constant. The particle’s position at t_0=0 is x_0=-7.90\:m. At t_1=1.00\:s, the particle is at x_1=8.70\:m.

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

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