part 1
A 1760 kg car accelerates uniformly from rest to 8.8 m/s in 4.52 s. Find the work done on the car in this time.
Answer in units of J.
part 2
Find the average power delivered by the engine in the first 1.808 s.
Answer in units of W.
part 3
Find the instantaneous power delivered by the engine at t2 = 2.7572 s.
Answer in units of W.
To find the answers to these questions, we'll need to use the formulas related to work, power, and acceleration. The equations we'll be using are:
1. Work (W) = Force (F) * Distance (d) * cos(theta)
2. Power (P) = Work (W) / Time (t)
3. Average Power (P_avg) = Change in Work (delta W) / Change in Time (delta t)
4. Instantaneous Power (P_inst) = dW / dt
Let's solve each part of the question step by step:
Part 1:
To find the work done on the car, we'll use the formula W = F * d * cos(theta). Since the car is accelerating uniformly, we can use the equation F = m * a, where m is the mass of the car and a is the acceleration. Given that the car starts from rest, its initial velocity (v0) is 0 m/s. The final velocity (v) is 8.8 m/s, and the time taken (t) is 4.52 s.
First, let's find the acceleration:
v = v0 + a * t
8.8 = 0 + a * 4.52
a = 8.8 / 4.52
Now, let's calculate the work done:
W = m * a * d * cos(theta)
Given the mass (m) of the car is 1760 kg and the distance (d) is not provided, we'll assume it to be the distance traveled during the given time. Since the car starts from rest, we can use the equation d = (v0 * t) + (0.5 * a * t^2):
d = (0 * 4.52 ) + (0.5 * (8.8 / 4.52) * (4.52)^2)
Now, we can substitute the values into the work formula to get the answer in joules (J).
Part 2:
To find the average power delivered by the engine in the first 1.808 s, we'll use the formula P_avg = delta W / delta t. The change in work (delta W) is simply the work done during the given time, and the change in time (delta t) is the given time period. We already have the work value from part 1. Now we can simply divide it by the given time to get the average power in watts (W).
Part 3:
To find the instantaneous power delivered by the engine at t2 = 2.7572 s, we'll use the formula P_inst = dW / dt, where dW is an infinitesimally small change in work and dt is an infinitesimally small change in time. Since the car is accelerating uniformly, we can assume constant power output throughout the acceleration. Therefore, the instantaneous power at any given time during the acceleration will be the same as the average power during that time. We can use the average power calculated in part 2 as the answer to part 3.