(1) The time taken for the drag racer to accelerate is 4.89 s
(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.
Formula:
- t = (v-u)/a........... Equation 1
Where:
- t = time
- v = Final velocity
- u = initial velocity
- a = acceleration.
From the question,
Given:
- a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²
- v = 60 mph = (60×0.44704) = 26.82 m/s
- u = 0 m/s ( from rest)
Substitute these values into equation 1
- t = (26.82-0)/5.4864
- t = 4.89 seconds
(2) To calculate the speed of the contestant, we use the formula below
Formula:
- s = d/t............ Equation 1
Where:
- s = speed of the contestant
- d = distance
- t = time.
From the question,
Given:
Substitute these values into equation 1
(a) In feet = (9.43/0.3048) = 30.94 ft/s
(b) in miles per hr = (9.43×2.24) = 21.12 miles per hr
Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
Learn more about speed here: brainly.com/question/4931057
