A car slows down from an initial velocity of 90.0 km/h to a final velocity of 36.0 km/h at a constant rate while undergoing a displacement of 105 m. (a) What is the acceleration of the car? (b) If the car continues with the same acceleration, how much additional time will it take to come to a stop? (c) What is the total displacement of the car from the instant when it's travelling at 90 km/h until it stops? (d) Sketch the position-time curve for the car on the empty graph on the next page. Indicate the time and position at which it stops.

The car moves at constant acceleration, so we can use kinematic equations, also known as SUVAT equations, to model the car's motion. For this problem, we will use the following equations:

v² = u² + 2as

v = u + at

where

s is the displacement
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time

(a) First, convert the speeds from km/h to m/s.

u = 90.0 km/h × (1000 m/km) × (1 h / 3600 s) = 25 m/s

v = 36.0 km/h × (1000 m/km) × (1 h / 3600 s) = 10 m/s

Given that s = 105 m, use the first equation to solve for the acceleration.

v² = u² + 2as

(10 m/s)² = (25 m/s)² + 2a (105 m)

a = -2.5 m/s²

(b) Given u = 10 m/s, v = 0 m/s, and a = -2.5 m/s², use the second equation to solve for time.

v = u + at

0 m/s = 10 m/s + (-2.5 m/s²) t

t = 4 s

(c) Given u = 25 m/s, v = 0 m/s, and a = -2.5 m/s², use the first equation to solve for the displacement.

v² = u² + 2as

(0 m/s)² = (25 m/s)² + 2 (-2.5 m/s²) s

s = 125 m

(d) Before we can graph the position-time curve, we need to solve for the total time. Using the second equation:

v = u + at

(0 m/s) = (25 m/s) + (-2.5 m/s²) t

t = 10 s

We now have 3 points we can use to construct the graph: (0, 0), (6, 105), and (10, 125). The car has constant acceleration, so the curve will be quadratic.

We can also use another kinematic equation to construct the curve:

s = ut + ½ at²

Given u = 25 and a = -2.5, the equation is:

s = 25t − 1.25t²

where 0 ≤ t ≤ 10.