Question

A proton in a synchrotron is moving in a circle of radius 1 km and increasing its speed by v(t)=c1+c2t², where c1=2.0×10⁵ m/s, c2=10⁵ m/s³.

(a) What is the proton’s total acceleration at t = 5.0 s?
a) 3.0 × 10⁵ m/s²
b) 4.0 × 10⁵ m/s²
c) 5.0 × 10⁵ m/s²
d) 6.0 × 10⁵ m/s²

Answer 1

The proton's total acceleration at t = 5.0 s is 10⁶ m/s².

Step-by-step explanation:

To find the proton's acceleration at t = 5.0 s, we need to differentiate the velocity function with respect to time, and then substitute t = 5.0 s into the resulting derivative. The velocity function is given as v(t) = c1 + c2t², where c1 = 2.0 × 10⁵ m/s and c2 = 10⁵ m/s³. The derivative of v(t) is dv(t)/dt = 2c2t.

Substituting t = 5.0 s into the derivative, we get dv(t)/dt = 2c2t = 2(10⁵ m/s³)(5.0 s) = 10⁶ m/s². Therefore, the proton's total acceleration at t = 5.0 s is 10⁶ m/s².