Question

A particles velocity is described by the function v(x)=kt^2 m/s, where k is a constant and t is in s. The particle's position at t0=0 is x0=-8.40m. At t1=2.00s, the particle is at x1=6.50 m. Determine the units of k in terms of m and s.

Answer 1

Step-by-step explanation:

Given

`v(x)=kt^2 m/s`

`At t_0=0 s, x_0=-8.40 m`

`At t_1=2 s x_1=6.50 m`

`\frac{\mathrm{d} x}{\mathrm{d} t}=v(x)=kt^2`

`x=(kt^3)/(3)+c`

`At\ t_0=0 s, x_0=-8.40 m`

-8.4=0+c

`x=(kt^3)/(3)-8.4`

At
`t_1=2 s x_1=6.50 m`

`6.5+8.4=(k* 2^3)/(3)`

`14.9* 3=k* 8`

`k=(44.7)/(8)=5.58 m/s^3`