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?
Vo = 80000m/3600s = 22.2 m/s.
V = 92000m/3600s = 25.6 m/s.
V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a.
To find the magnitude of the car's displacement, we need to first convert the speeds from km/h to m/s.
Given:
Initial speed (u) = 80 km/h
Final speed (v) = 92 km/h
Acceleration (a) = 0.83 m/s^2
Step 1: Convert the speeds from km/h to m/s.
To convert km/h to m/s, we divide by 3.6 since there are 3.6 seconds in an hour.
Initial speed in m/s: u = 80 km/h / 3.6 = 22.22 m/s (rounded to two decimal places)
Final speed in m/s: v = 92 km/h / 3.6 = 25.56 m/s (rounded to two decimal places)
Step 2: Calculate the displacement using the equation:
displacement (s) = (v^2 - u^2) / (2a)
Plug in the values:
s = (25.56^2 - 22.22^2) / (2 * 0.83)
(Note: ^ denotes exponentiation)
Calculating the numerator:
25.56^2 = 654.12
22.22^2 = 492.48
Subtracting the numerator:
654.12 - 492.48 = 161.64
Dividing the numerator by the denominator:
s = 161.64 / (2 * 0.83)
Calculating the denominator:
2 * 0.83 = 1.66
Dividing the numerator by the denominator:
s = 97.56 m (rounded to two decimal places)
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 97.56 m.