Answer to Question #235575 in Mechanics | Relativity for Pankhss

Question

A particle travels along a straight line with a constant acceleration. When, s = 4 ft, v = 3 ft/s and when

s = 10 ft, v = 8 ft/s. Determine the velocity as a function of position.

Answer 1

The traveled distance is

$S=S_2-S_1 \\ S=10-4 \\ S=6 \;ft$

Calculate the acceleration using the following formula:

$v^2_1 -v^2_2 = 2as \\ 8^2 -3^2 = 2 \times a \times 6 \\ a=\frac{55}{12} \\ a= 4.58 \;ft/s^2$

Calculate the velocity using the following formula:

$a=\frac{dv}{dt} \\ a=(\frac{dv}{ds})(\frac{ds}{dt}) \\ a= v \times (\frac{dv}{ds}) \\ v \times dv = 4.58 \times ds$

Apply the integration on both sides.

$\int vdv = 4.58 \int ds \\ \frac{v^2}{2}=4.58 \times s +C \\ \frac{3^2}{2}= 4.58 \times 4 + C \\ C= -13.82 \\ \frac{v^2}{2}=4.58 \times s -13.82 \\ v^2 = 9.16s -27.6 \\ v = \sqrt{(9.16s-27.6)}$