Final answer:
It will take approximately 3.8764 × 10^-4 seconds for the car to reach a speed of 11.0 m/s, neglecting friction. If the car also climbs a 2.80 m-high hill in the process, it will take approximately 0.8061 seconds.
Step-by-step explanation:
To find the time it takes for the car to reach a speed of 11.0 m/s, we can use the equation:
Power = Force × Velocity
Given that Power = 38.0 hp and Velocity = 11.0 m/s, we need to convert the power from horsepower to watts:
38.0 hp × 746 W/hp = 28348 W
Now we can rearrange the equation to solve for time:
Time = Force ÷ (Mass × Acceleration)
Since the car is starting from rest, the initial velocity is 0 m/s. The final velocity is 11.0 m/s. Rearranging the equation to solve for acceleration:
Acceleration = (Final Velocity - Initial Velocity) ÷ Time
Substituting the values into the equation:
(11.0 m/s - 0 m/s) ÷ Time = 2.8348 × 10^4 N
Simplifying the equation:
11.0 m/s ÷ Time = 2.8348 × 10^4 N
Time = 11.0 m/s ÷ (2.8348 × 10^4 N)
Calculating the time:
Time = 3.8764 × 10^-4 s
Therefore, it will take approximately 3.8764 × 10^-4 seconds for the car to reach a speed of 11.0 m/s, neglecting friction.
If the car also climbs a 2.80 m-high hill in the process, we need to consider the work done against gravity by the car. The equation for work done is:
Work = Force × Distance
Given that the car climbs a 2.80 m-high hill, we can calculate the work done:
Work = Mg × h
where M is the mass of the car (870 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill (2.80 m).
Substituting the values into the equation:
Work = 870 kg × 9.8 m/s^2 × 2.80 m
Calculating the work:
Work = 22867.2 Joules
Now we can find the time it takes to climb the hill by dividing the work done by the power output of the car:
Time = Work ÷ Power
Substituting the values into the equation:
Time = 22867.2 Joules ÷ (28348 W)
Calculating the time:
Time = 0.8061 s
Therefore, it will take approximately 0.8061 seconds for the car to reach a speed of 11.0 m/s and climb a 2.80 m-high hill.