A certain car can accelerate from rest to the speed v in time T seconds.
(a) If the power output of the car remains constant, how long does it take for the car to accelerate from v to 3v, in terms of T?
(b) How fast is the car moving 9T after starting, in terms of v?
(a) Well, if the car can accelerate from rest to the speed v in time T, I guess we can say it takes T units of time for the car to accelerate by one v. So, if we want to go from v to 3v, we just need to accelerate by two v's. Therefore, it would take 2T units of time.
(b) After starting, the car would be moving 9T units of time later. Now, if it takes T units of time to accelerate from rest to v, and in 9T units of time the car has gone through 9 times the acceleration, then it would be moving at 9v. So the car would be moving at 9v. In other words, it really picked up some speed!
To answer these questions, we need to use some principles of constant power and uniform acceleration.
Let's start by defining the variables:
- T: Time taken for the car to accelerate from rest to speed v (in seconds)
- v: The initial speed of the car (in any consistent unit)
- 3v: The final speed of the car (in any consistent unit)
- a: Acceleration of the car (in meters per second squared)
- P: Power output of the car (in watts)
- d: Distance traveled during acceleration (in meters)
(a) If the power output of the car remains constant, we can use the equation for power, P = (1/2) m v^2 / T, where m is the mass of the car. Since P is constant, we can write:
(1/2) m v^2 / T = (1/2) m (3v)^2 / t
Simplifying, we get:
v^2 / T = 9v^2 / T
Cancelling out v^2 and T on both sides, we find:
1 / T = 9 / t
Therefore, the time required for the car to accelerate from v to 3v is 9T seconds.
(b) To determine how fast the car is moving 9T seconds after starting, we need to calculate the distance traveled during that time.
Using the equation for uniformly accelerated motion, d = (1/2) a t^2, we can find the distance traveled by the car in T seconds:
d = (1/2) a T^2
Since the power output of the car remains constant, the acceleration of the car is the same throughout. Therefore, the distance traveled in 9T seconds will be:
9d = (1/2) a (9T)^2 = (1/2) a 81T^2
Now, using the equation for speed, v = d / t, we can calculate the speed of the car 9T seconds after starting:
v = (1/2) a 81T^2 / (9T) = (1/2) a 9T
The speed of the car 9T seconds after starting, in terms of v, is (1/2) a 9T.
To answer these questions, we need to understand the relationship between power, acceleration, and time during an acceleration.
(a) If the power output of the car remains constant, it means that the acceleration also remains constant. Let's call the initial acceleration of the car "a" and the initial speed "v". According to the equation of motion, we know that:
v = a * T (Equation 1)
Now we need to find the time it takes for the car to accelerate from v to 3v. Let's call this time "t".
To find t, we need to calculate the distance covered during this time. The distance covered can be expressed as the area under the velocity-time graph.
The velocity of the car is changing uniformly from v to 3v over time t. So the average velocity during this time period is (v + 3v) / 2 = 2v.
Since the acceleration remains constant, we can calculate the average velocity using the formula:
Average velocity = Initial velocity + (Final velocity - Initial velocity) / 2
Therefore:
2v = v + a * t / 2
Substituting the value of "v" from Equation 1, we get:
2aT = a * t / 2
2T = t / 2
t = 4T
So, it takes 4 times the initial time T for the car to accelerate from v to 3v.
(b) To find how fast the car is moving 9T after starting, we need to calculate the final speed at that time. Let's call the final speed "v_prime".
Using Equation 1, we know that:
v = a * T
So, the speed after time 9T can be calculated as:
v_prime = a * (9T) = 9aT
Therefore, the car is moving at a speed of 9v after 9T seconds.
In summary:
(a) It takes 4 times the initial time T for the car to accelerate from v to 3v.
(b) The car is moving at a speed of 9v after 9T seconds.
v=a•t
a=v/t
If the power output of the car remains constant,a=const
v(f) =v(i) +a•t1=>
3•v=v+v•t1/t
2•v=v•t1/t
t1=2t
(b) v2=a•9•t= v•9•t/t=9•v