Answer to Question #198353 in Mechanics | Relativity for Mohau Mathews Mell

Question

The velocity of an object is given by the expression v(t) = 3.00 m/s + (4.00 m/s³)t², where t is in seconds. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s .

Answer 1

velocity of an object is given by the expression

$v(t) = 3.00 \;m/s + (3.00\; m/s^3)t^2$

located at x = 1.00 m

at time t = 0.000

Find the position of the object as a function of time.

Now,

$v(t) = 3.00 \;m/s + (3.00 \;m/s^3)t^2 \\ v=\frac{dx}{dt}=3.00 \;m/s + (3.00 \;m/s^3)t^2 \\ \int dx=\int (3.00 m/s + (3.00\; m/s^3)t^2 )dt \\ x=3t+\frac{3t^3}{3}+C \\ t=0 \\ x=1 \\ 1 = 3 \times 0 + \frac{3 \times 0^3}{3} + C \\ C=1 \\ x=3t+\frac{3t^3}{3}+1$

The position of the object as a function of time

$x=t^3+3t+1$