1. What constant acceleration, in SI units, must a car have to go from zero to 69.0 mph in 13.0s ?
2. What fraction of g is this? (the unit for this answer is % of g)
Wouldn't it help to change 69mph to m/s?
sds
To find the constant acceleration needed for the car to go from zero to 69.0 mph in 13.0 seconds, we can use the following steps:
Step 1: Convert the speed from miles per hour to meters per second (SI unit).
1 mile = 1.60934 kilometers
1 kilometer = 1000 meters
1 hour = 3600 seconds
69.0 mph = 69.0 * 1.60934 * 1000 / 3600 m/s
≈ 30.83 m/s
Step 2: Use the formula for constant acceleration:
v = u + at
where:
v = final velocity (30.83 m/s)
u = initial velocity (0 m/s)
a = acceleration (unknown)
t = time (13.0 seconds)
Rearranging the equation:
a = (v - u) / t
a = (30.83 m/s - 0 m/s) / 13.0 s
a ≈ 2.37 m/s^2
Therefore, the car must have a constant acceleration of approximately 2.37 m/s^2 to go from zero to 69.0 mph in 13.0 seconds.
To find the fraction of g this acceleration represents, we can compare it to the acceleration due to gravity, denoted as "g." On Earth's surface, the average value of acceleration due to gravity is approximately 9.8 m/s^2.
Step 1: Calculate the fraction of g.
Fraction of g = (acceleration / g) * 100%
Fraction of g = (2.37 m/s^2 / 9.8 m/s^2) * 100%
Fraction of g ≈ 24.18%
Therefore, the acceleration required for the car is approximately 24.18% of the acceleration due to gravity (g).