A certain car is capable of accelerating at a

uniform rate of 0.83 m/s2
.
What is the magnitude of the car’s displacement
as it accelerates uniformly from a speed
of 80 km/h to one of 92 km/h?
Answer in units of m.

Use

v1^2-v0^2=2(a)(Δx)
or
Δx = (v1^2-v0^2)/(2a)
a=acceleration,
v0=initial velocity
v1=final velocity

Do not forget to convert km/h to m/s to get your answer in metres.

To find the magnitude of the car's displacement, we need to determine the distance it travels during the acceleration.

First, let's convert the initial and final speeds from km/h to m/s.

Initial Speed (v1) = 80 km/h
Final Speed (v2) = 92 km/h

1 km/h = 0.27778 m/s
So, v1 = 80 km/h * 0.27778 m/s = 22.222 m/s
and v2 = 92 km/h * 0.27778 m/s = 25.556 m/s

Next, let's calculate the time it takes for the car to accelerate from v1 to v2. We'll use the formula:

v2 = v1 + at

where v2 is the final speed (25.556 m/s), v1 is the initial speed (22.222 m/s), a is the acceleration (0.83 m/s^2), and t is the time. Rearranging the formula, we have:

t = (v2 - v1) / a

t = (25.556 m/s - 22.222 m/s) / 0.83 m/s^2
t = 3.334 m/s / 0.83 m/s^2
t = 4 seconds (approximately)

Now, we can find the displacement using the formula:

s = v1t + (1/2)at^2

where v1 is the initial speed (22.222 m/s), t is the time (4 seconds), and a is the acceleration (0.83 m/s^2).

s = (22.222 m/s * 4 s) + (0.5 * 0.83 m/s^2 * (4 s)^2)
s = 88.888 m + 6.664 m
s = 95.552 m

Therefore, the magnitude of the car's displacement as it accelerates uniformly from a speed of 80 km/h to one of 92 km/h is 95.552 meters.