The engine in an imaginary sports car can provide constant power to the wheels o
ID: 1261503 • Letter: T
Question
The 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 31.0mph in time 1.50s .
Part A
At full power, how long would it take for the car to accelerate from 0 to 62.0mph ? Neglect friction and air resistance.
Express your answer in seconds.
Part 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 31.0mph in time 1.5s , how long would it take to go from zero to 62.0mph ?
Explanation / Answer
A)
Power = Work done/ time
As power here is constant,so, work done in every second is same
Also, Work done = change in Kinetic energy
So, change in Kinetic energy every second is same in this case
But change in Kinetic energy is directly proportional to change in square of velocity.
So, for the engine to speed up from 0 to 31 mph (or 31*1.61/3.6 = 13.86 m/s),
change in square of velocity = (13.86^2-0^2) = 192.2 m2/s2
Now, this change is in 1.5 s
So, in 1 second, the change will be 192.2/1.5 = 128.14 m2/s2
Now, for the final speed to be 62 mph ( or 62*1.61/3.6 = 27.73 m/s) ,
the change in square of speed = 27.73^2-0^2 = 768.83 m2/s2
But in 1 second it can change the square of speed by 128.14 m2/s2
So, time taken for 768.83 m2/s2 change = 768.83/128.14 = 6 s <---------answer
B)
initial speed, u = 0
final speed, v = 31 mph = 31*1.61/3.6 m/s = 13.86 m/s
time taken, t = 1.5 s
So, acceleration, a = (v-u)/t = (13.86-0)/1.5 = 9.24 m/s2
So, for, v = 62 mph = 62*1.61/3.6 m/s = 27.73 m/s,
From the equation, a = (v-u)/t
So, t = (v-u)/a = (27.73-0)/9.24 = 3 s <-----------answer
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