Answer:
a. 1.93 s b. 3 s
Explanation:
a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?
Since the car accelerates from 0 to 30 mph in 1.00 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 30 mph and t = 1.00s = 1/3600 h
So, substituting the values of the variables into the equation, we have
a = (v - u)/t
a = (30 mph - 0 mph)/ 1/3600 h
a = 30 mph × 3600 /h
a = 108000 mph²
So, we find the time it takes the car to accelerate to 58 mph from 0 mph from
t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 108000 mph²
So, substituting the value of the variables into the equation, we have
t = (v' - u')/a
t = (58 mph - 0 mph)/108000 mph²
t = 58 mph/108000 mph²
t = 5.37 × 10⁻⁴ h
t = 5.37 × 10⁻⁴ × 3600 s
t = 1.93 s
b. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0mph in time 1.50s , how long would it take to go from zero to 58.0mph ?
Since the car accelerates from 0 to 29 mph in 1.50 s, we find its acceleration, a from a = (v - u)/t where u = 0 m/s, v = 29 mph and t = 1.05s = 1/3600 h
So, substituting the values of the variables into the equation, we have
a = (v - u)/t
a = (29 mph - 0 mph)/ 1.5/3600 h
a = 29 mph × 3600/1.5 /h
a = 104400/1.5 mph²
a = 69600 mph²
So, we find the time it takes the car to accelerate to 58 mph from 0 mph from
t = (v' - u')/a where u = 0 mph, v = 58 mph and a = 69600 mph²
So, substituting the value of the variables into the equation, we have
t = (v' - u')/a
t = (58 mph - 0 mph)/69600 mph²
t = 58 mph/69600 mph²
t = 8.33 × 10⁻⁴ h
t = 8.33 × 10⁻⁴ × 3600 s
t = 3 s
