A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.40 m/s2 for 14.0 s.
2. Maintain a constant velocity for the next 2.85 min.
3. Apply a constant negative acceleration of −8.39 m/s2 for 4.00 s.
(a) What was the total displacement for the trip?
_______m
(b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?
leg 1. ________ m/s
leg 2. ________ m/s
leg 3. ________ m/s
complete trip ________ m/s
I really need help with this question. I keep getting it wrong and im struggling with it.
To solve this problem, we can break it down into three parts: the first part with constant acceleration, the second part with constant velocity, and the third part with constant negative acceleration.
First, let's find the total displacement for each part:
For part 1:
Using the equation of motion, d = v₀t + (1/2)at², where d is the displacement, v₀ is the initial velocity, a is the acceleration, and t is the time:
d₁ = (0)(14.0) + (1/2)(2.40)(14.0)²
d₁ = 0 + (1/2)(2.40)(196.00)
d₁ = (1.20)(196.00)
d₁ ≈ 235.20 m
For part 2:
The displacement for this part is simply the product of the constant velocity and the time:
d₂ = (V)(t)
where V is the constant velocity and t is the time:
d₂ = (V)(2.85 min)(60 s/min)
Convert the time to seconds:
d₂ = (V)(171.00 s)
We are not given the value of V, so we will leave it as V for now.
For part 3:
Using the equation of motion, d = v₀t + (1/2)at²:
d₃ = (-V')(4.00) + (1/2)(-8.39)(4.00)²
d₃ = (-V')(4.00) + (1/2)(-8.39)(16.00)
d₃ = (-V')(4.00) + (-67.04)
d₃ = (-4.00)(V') - 67.04
where V' is the initial velocity for part 3 (which is equal to V from part 2), and we will leave it as V' for now.
(a) To find the total displacement for the trip, we need to sum up the displacements for each part:
Total displacement = d₁ + d₂ + d₃
Total displacement = 235.20 + V + (-4.00V') - 67.04
(b) To find the average speed for each part, we need to divide the total displacement for each part by the time taken for that part:
Average speed for leg 1:
v₁ = d₁/t₁
where t₁ is the time taken for part 1, which is given as 14.0 s
Average speed for leg 2:
v₂ = d₂/t₂
where t₂ is the time taken for part 2, which is given as 2.85 min (convert to seconds)
Average speed for leg 3:
v₃ = d₃/t₃
where t₃ is the time taken for part 3, which is given as 4.00 s
Average speed for the complete trip:
v_complete = (Total displacement)/Total time
where Total time is the sum of t₁, t₂, and t₃ in seconds.
To obtain the numerical values, we need the values of V and V' from part 2 to substitute into our equations. Please provide the values of V and V'.
To find the total displacement for the trip, we need to determine the individual displacements for each leg and then sum them up.
For the first leg:
We are given the constant acceleration (a) of 2.40 m/s^2 and the time (t) of 14.0 seconds. We can use the following equation to find the displacement (s):
s = ut + (1/2)at^2
Starting from rest, the initial velocity (u) is 0 m/s.
s = 0 * 14.0 + (1/2) * 2.40 * (14.0)^2
s = 0 + 168.56
s = 168.56 m
So, the displacement for the first leg is 168.56 meters.
For the second leg:
We are told to maintain a constant velocity, which means there is no acceleration. Therefore, the displacement for this leg is simply the product of the velocity (v) and the time (t):
s = v * t
Given that the time is 2.85 minutes, we need to convert it to seconds:
t = 2.85 min * 60 s/min
t = 171.00 s
We are not given the velocity, but since it is constant, we can use the final velocity from the first leg (which is also the initial velocity for this leg) since there is no acceleration or deceleration. Therefore, the displacement for this leg is:
s = 2.40 * 14.00 + (1/2) * 0 * (171.00 - 14.00)^2
s = 33.60 + 0
s = 33.60 m
So, the displacement for the second leg is 33.60 meters.
For the third leg:
We are given the constant negative acceleration (a) of -8.39 m/s^2 and the time (t) of 4.00 seconds. Using the same formula as before:
s = ut + (1/2)at^2
The initial velocity for this leg is the final velocity from the second leg (since there is no change in velocity between the two legs). Therefore, the final velocity from the second leg is:
v = u + at
v = 2.40 + (-8.39) * 4.00
v = 2.40 - 33.56
v = -31.16 m/s
Now, we can find the displacement for the third leg:
s = -31.16 * 4.00 + (1/2) * (-8.39) * (4.00)^2
s = -124.64 + (-66.80)
s = -191.44 m
Note that the displacement is negative because the direction is opposite to the initial direction.
So, the displacement for the third leg is -191.44 meters.
Now, to find the total displacement for the trip, we simply add up the individual displacements:
Total displacement = displacement of leg 1 + displacement of leg 2 + displacement of leg 3
Total displacement = 168.56 + 33.60 + (-191.44)
Total displacement = 10.72 meters
Therefore, the total displacement for the trip is 10.72 meters.
Now, to find the average speeds for each leg and the complete trip, we need to use the formula for average speed:
Average speed = total distance / total time
For leg 1:
The distance is equal to the displacement since the direction is the same.
Average speed of leg 1 = 168.56 m / 14.00 s
Average speed of leg 1 = 12.04 m/s (rounded to two decimal places)
For leg 2:
The distance traveled is equal to the displacement since the direction is the same.
Average speed of leg 2 = 33.60 m / 171.00 s
Average speed of leg 2 = 0.20 m/s (rounded to two decimal places)
For leg 3:
The distance is equal to the magnitude of the displacement.
Average speed of leg 3 = 191.44 m / 4.00 s
Average speed of leg 3 = 47.86 m/s (rounded to two decimal places)
For the complete trip:
The total distance is equal to the total displacement.
Average speed of the complete trip = 10.72 m / (14.00 + 171.00 + 4.00) s
Average speed of the complete trip = 0.04 m/s (rounded to two decimal places)
Therefore, the average speeds for legs 1, 2, and 3 of the trip are:
Average speed of leg 1 = 12.04 m/s
Average speed of leg 2 = 0.20 m/s
Average speed of leg 3 = 47.86 m/s
Average speed of the complete trip = 0.04 m/s