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

The velocity of the electron at different time intervals are: A) 0 m/s, B) -2.3 m/s, C) -12.2 m/s, D) -34.7 m/s.

Step-by-step explanation:

To determine the velocity at different times, we need to take the derivative of the position function with respect to time. The velocity is the derivative of the position function, so its formula is:

v(t) = 3pt² + 2qt

For part A, when t=0, the velocity formula gives us:

v(0) = 3(0)(0)² + 2(1.5)(0) = 0 m/s

For part B, when t=1s, the velocity formula gives us:

v(1) = 3(-2.3)(1)² + 2(1.5)(1) = -2.3 m/s

For part C, when t=2s, the velocity formula gives us:

v(2) = 3(-2.3)(2)² + 2(1.5)(2) = -12.2 m/s

For part D, when t=3s, the velocity formula gives us:

v(3) = 3(-2.3)(3)² + 2(1.5)(3) = -34.7 m/s

Answer 2

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

Step-by-step 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`