A proton (mass m = 1.67 multiplied by 10-27 kg) is being accelerated along a straight line at 2.2 multiplied by 1015 m/s2 in a machine. The proton has an initial speed of 2.4 multiplied by 107 m/s and travels 3.8 cm.

(a) What is its speed?
m/s
(b) What is the increase in its kinetic energy? j

a. V^2 = Vo^2 + 2ad,

V^2 = (2.4*10^7)^2 + 4.4*10^15*0.038,
V^2=5.76*10^14+1.67*10^14=7.43*10^14,
V = 2.73*10^7.

b. KE = 0.5m*V^2,
KE1=(1.67/2)(2.4*10^7)^2=4.81*10^14J.
KE2=(1.67/2)*(2.73*10^7)^2=6.22*10^14J.

Increase = KE2 - KE! =
6.22*10^14 - 4.81*10^14 = 1.41*10^14J.

To solve this problem, we need to apply the equations of motion. Using the equation:

Vf^2 = Vi^2 + 2ad

Where:
- Vf is the final velocity
- Vi is the initial velocity
- a is the acceleration
- d is the distance

(a) First, let's find the final velocity of the proton. We'll substitute the given values into the equation:

Vf^2 = (2.4x10^7 m/s)^2 + 2(2.2x10^15 m/s^2)(0.038 m)

Now we can calculate the final velocity:

Vf = √((2.4x10^7 m/s)^2 + 2(2.2x10^15 m/s^2)(0.038 m))

(b) To calculate the increase in kinetic energy, we need to use the equation:

ΔKE = 1/2 mvf^2 - 1/2 mvi^2

Where:
- ΔKE is the change in kinetic energy
- m is the mass
- vf is the final velocity
- vi is the initial velocity

Now we can substitute the values into the equation and calculate the change in kinetic energy:

ΔKE = 1/2 (1.67x10^-27 kg) (Vf^2 - (2.4x10^7 m/s)^2)

Solving this equation will give us the increase in kinetic energy in joules (J).