The engine in an imaginary sportThe engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 30.0 mph in time 1.00 s .s car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). Required:
a. At full power, how long would it take for the car to accelerate from 0 to 58.0mph?
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 ?

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

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