A certain car is capable of accelerating at a uniform rate of 0.93 m/s2. What is the magnitude of the car’s displacement as it accelerates uniformly from a speed of 86 km/h to one of 98 km/h?

Answer in units of m

vf^2=vi^2+2ad solve for d.

be certain to change km/hr to m/s...

v = Vi + a t

Vi = 86 km/h * 1000 m/km * 1 h/3600 s
= 23.9 m/s

Vf = 98 km/h *10/36 = 27.2 m/s

hard way:
27.2 = 23.9 + .93 t
so
t = 3.55 s

x = Xi + Vi t + (1/2) a t^2
x-Xi = displacement = 23.9(3.55)+.465(3.55)^2
= 90.7 m

easy way:
average speed = (23.9+27.2)/2 = 25.55
change in speed/t = .93 = 3.3/t
t = 3.55 sure enough
distance = 25.55*3.55 = 90.7 m

To find the displacement of the car, we need to calculate the change in distance traveled. We know that the car accelerates uniformly, so we can use the equation:

v^2 = u^2 + 2as

Where:
- v is the final velocity of the car (98 km/h)
- u is the initial velocity of the car (86 km/h)
- a is the acceleration of the car (0.93 m/s^2)
- s is the displacement

First, we need to convert the velocities from km/h to m/s by dividing by 3.6 (since 1 km/h = 1/3.6 m/s).

Final velocity (v) = 98 km/h = (98/3.6) m/s ≈ 27.22 m/s
Initial velocity (u) = 86 km/h = (86/3.6) m/s ≈ 23.89 m/s

Now, we can solve for the displacement (s):

s = (v^2 - u^2) / (2a)
= (27.22^2 - 23.89^2) / (2 * 0.93)
≈ 2.225 meters (rounded to 3 decimal places)

Therefore, the magnitude of the car's displacement as it accelerates uniformly from a speed of 86 km/h to one of 98 km/h is approximately 2.225 meters.