Question about positions and velocity?

Acceleration is not constant velocity.

Acceleration is change in velocity per unit time (d)V/(d)t

80ft/s ÷ 4s = 20ft/s²

Distance = ½at²

D = ½×20ft/s²×(4s)² = 160 ft

Part a you got wrong, you're right that it should be a negative acceleration, but the value and units are wrong.

Acceleration is metres/seconds^2. Or feet or inches or whatever you want to use.

If the acceleration is -80ft/s^2 it would only take one second to lose that 80ft/s of initial velocity. The question states that it takes 4 seconds rather than one, so the acceleration must be 4x as weak, that is 20ft/s^2.

The answer to part b is the displacement of the car which would be the anti derivative if it's velocity which can be modelled as 80 - 20t = v

You could anti derive that and check the position at t=4 which turn out to be:

80t - 10t^2

= 80(4)-10(4^2)

=320 - 160 = 160ft

A) Acceleration = change in velocity / time taken for the change

Here the acceleration [decelleration] is constant, so you don't need calculus.

Change in velocity = final velocity - initial velocity = 0 - 80 = - 80 m/s

Time taken = 4

Acceleration = - 80 / 4 = - 20 m/s^2

B. Recall that

v^2 = u^2 - 2as

v = final velocity (m/s); u = initial velocity (m/s); a = acceleration (m/s^2); s = distance (m)

s = (u^2 - v^2) / 2a

s = 80^2 - 0^2 / 2(20) = 6400 / 40 = 160 m

As a tip. You only need to use calculus for equations where the accelerations or velocities are NOT constant. If they are, you just need the regular equations of motion.

We assume CONSTANT acceleration

(A) a = (0 - 80) / 4 = -20 ft/s^2

(B) Vf^2 = Vi^2 + 2ad, ====> d = 160 ft