Question

The motion of an electron is given by x(t)=pt3+qt2+r, with p = -2.3 m/s^3 ,q = +1.5 m/s^2 , and r = +9.0 m.A) Determine its velocity at t=0.B) Determine its velocity at t=1s.C) Determine its velocity at t=2s.D) Determine its velocity at t=3s.

Answer 1

Answer:

$v(0)=0\\v(1)=-3.9\ m/s\\v(2)=-21.6\ m/s\\v(3)=-53.1\ m/s$

Explanation:

Instant Velocity

Given the position as a function of time x(t), the instant velocity is the derivative of the function:

$v(t)=x'(t)$

We are given the position as

$x(t)=-2.3t^3+1.5t^2+9$

The derivative of x is

$v(t)=x'(t)=-6.9t^2+3.0t$

A) Let's compute v(0)

$v(0)=-6.9(0)^2+3.0(0)=0$

B)  

$v(1)=-6.9(1)^2+3.0(1)$

$v(1)=-3.9\ m/s$

C)

$v(2)=-6.9(2)^2+3.0(2)$

$v(2)=-21.6\ m/s$

D)

$v(3)=-6.9(3)^2+3.0(3)$

$v(3)=-53.1\ m/s$