Question

A proton in a synchrotron is moving in a circle of radius 1 km and increasing its speed by v(t) = c1 + c2 t 2 , where c1 = 2.0 × 105m/s, c2 = 105m/s3 .

(a) What is the proton’s total acceleration at t = 5.0 s?
(b) At what time does the expression for the velocity become unphysical?

Answer 1

a) 10^6 m/s^2

b) 7.39 s

Step-by-step explanation:

Hello!

If the velocity is given by:

`v(t) = c_1 + c_2 t^2`

Then the acceleration is the derivative of the velocity with respect of time:

`a(t) = 2 c_2 t`

a)

Therefore the acceleration at t=5.0s is:

`a(5.0s) = 2 c_2 * 5.0 = 10^6 m/s^2`

b)

The velocity will become unphysical when it equals the velocity of ligth (3x10^8 m/s)

3x10^8 m/s = 2x10^5 m/s + 10^5 m/s^3 t^2

Solving for t:

t^2 = 2998 s^2

t = 7.39 s