A record of travel along a straight path is as
follows:
(a) Start from rest with constant acceleration of 3.07 m/s2 for 16.5 s;
(b) Constant velocity of 50.655 m/s for the next 0.767 min;
(c) Constant negative acceleration of
−9.49 m/s2 for 5.89 s.
What was the total displacement x for the complete trip?
Well, this sounds like a rollercoaster ride! Let's calculate the total displacement.
First, let's find the displacement during the first part of the journey.
(a) To find the displacement during the first part, we can use the formula s = ut + (1/2)at^2.
The initial velocity (u) is 0 since you started from rest, the acceleration (a) is 3.07 m/s^2, and the time (t) is 16.5 seconds.
Using the formula, we get:
s1 = (0)(16.5) + (1/2)(3.07)(16.5)^2
Now, let's calculate the displacement during the second part.
(b) During the second part, you traveled at a constant velocity of 50.655 m/s for 0.767 minutes.
Since velocity is constant, the displacement is given by s = vt.
Using the formula, we get:
s2 = (50.655)(0.767)(60)
Finally, let's calculate the displacement during the third part.
(c) During the third part, you experienced a constant negative acceleration of -9.49 m/s^2 for 5.89 seconds.
Using the formula, we get:
s3 = (50.655)(5.89) + (1/2)(-9.49)(5.89)^2
Now, let's add up all the displacements:
Total displacement (x) = s1 + s2 + s3
I'll let you do the math, but be careful not to drop any numbers along the way!
To find the total displacement for the complete trip, we need to find the displacement for each segment of the trip and then add them together.
(a) Using the equation of motion:
v = u + at
where
v = final velocity,
u = initial velocity,
a = acceleration,
t = time,
we can find the final velocity at the end of the first segment:
v = u + at
v = 0 + 3.07 * 16.5
v = 50.655 m/s
The displacement during the first segment can be found using the formula:
s = ut + (1/2)at^2
where
s = displacement,
Since the initial velocity is 0, the equation simplifies to:
s = (1/2)at^2
s = (1/2) * 3.07 * (16.5)^2
s ≈ 265.758 m
(b) The next segment has constant velocity, so the displacement is given by:
s = v * t
s = 50.655 * 0.767 * 60
s ≈ 2331.23 m
(c) The final segment has constant negative acceleration, so we can use the same equation as in the first segment:
v = u + at
Since the final velocity is 0, the equation becomes:
0 = u + at
0 = 50.655 + (-9.49) * 5.89
u ≈ -245.604 m/s
Using the same equation to find the displacement:
s = ut + (1/2)at^2
s = -245.604 * 5.89 + (1/2) * (-9.49) * (5.89)^2
s ≈ -891.951 m
To find the total displacement, we add the displacements for each segment:
Total displacement = 265.758 + 2331.23 - 891.951
Total displacement ≈ 1705.037 m
Therefore, the total displacement for the complete trip is approximately 1705.037 meters.
To find the total displacement for the complete trip, we need to calculate the displacements for each segment of motion and then sum them up.
Let's break down the given information and calculate the displacement for each segment:
(a) Start from rest with constant acceleration of 3.07 m/s² for 16.5 seconds:
In this segment, we can use the kinematic equation:
displacement (s) = initial velocity (v₀) * time (t) + 0.5 * acceleration (a) * time² (t²)
Given that the initial velocity is 0 (start from rest), acceleration is 3.07 m/s², and time is 16.5 seconds:
s₁ = 0 * 16.5 + (0.5 * 3.07 * (16.5)²)
Simplifying the equation:
s₁ = 0 + 0.5 * 3.07 * 272.25
s₁ = 0 + 418.39325
s₁ ≈ 418.4 m
(b) Constant velocity of 50.655 m/s for the next 0.767 minutes:
Since the velocity is constant, the displacement can be calculated using:
displacement (s) = velocity (v) * time (t)
Given that the velocity is 50.655 m/s and time is 0.767 minutes:
s₂ = 50.655 * (0.767 * 60)
Converting minutes to seconds:
s₂ = 50.655 * 46.02
s₂ ≈ 2329.8 m
(c) Constant negative acceleration of −9.49 m/s² for 5.89 seconds:
Using the same kinematic equation as in (a), but with a negative acceleration value:
s₃ = 50.655 * 5.89 + (0.5 * -9.49 * (5.89)²)
s₃ = 298.17 - 166.71539
s₃ ≈ 131.45 m
Now, we can calculate the total displacement by summing up the individual displacements:
Total displacement, x = s₁ + s₂ + s₃
x ≈ 418.4 m + 2329.8 m + 131.45 m
x ≈ 2879.65 m
Therefore, the total displacement for the complete trip is approximately 2879.65 meters.