3 part physics question?

Question

(Part 1)
A dragster and driver together have mass
885.3 kg. The dragster, starting from rest,
attains a speed of 26.9 m/s in 0.58 s.
Find the average acceleration of the drag-
ster during this time interval. Answer in units
of m/s^2.

(Part 2)
What is the size of the average force on the
dragster during this time interval? Answer in
units of N.

(Part 3)
Assume: The driver has a mass of 79 kg.
What horizontal force does the seat exert
on the driver? Answer in units of N.

Answer 1

Part 1. v = at
26.9 = 0.58a
a = 46.379 m/s^2 (round off to 46.4)

Note that the dragster and the driver are both accelerating at the same rate.

Part 2. F = ma = 885.3 x 46.379 = 41059.60 N (round off to 41060 N)

What's going on here is that the engine, transmission, axle, etc. are transmitting power to the wheels. The rotating wheels, together with the coefficient of friction, exert a horizontal force against the pavement, and for every action, there's an equal and opposite reaction. So the pavement pushes back, and the vehicle moves ahead. This reactive force propels the combined driver/dragster 885.3 kg system.

Part 3. F = ma = 79 x 46.379 = 3663.97 N (round off to 3664)

Here, the seat is pushing the driver forward.

Answer 2

Part 1

acceleration = v(final) - v(intial) / time

i'm hoping you can do the rest yourself? Just plug in numbers
v(intial) is zero

Part 2

F = MA

mass of dragster, acceleration of dragster from part 1
Again, plug in the numbers yourself

Part 3

I'm not sure on part 3, I'd assume F = MA again
but this time you use mass of driver
No promises

Answer 3

distance equalst to sq. of preliminary velocity plus 0.5 of the value of acceleration more advantageous via the sq. of time this formulation saves physics student's lifes in tests. Rule of physic selection a million