A motor cycle, starting from rest, has an acceleration of +2.6 m/s^2. When the motorcycle has traveled a distance of 120 m, it slows down with an acceleration of -1.5 m/s^2 until its velocity is +12 m/s. What is the total displacement of the motorcycle?
There are two motions of the journey. The first motion's displacement is given: 120m
Now you calculate the 'Final Velocity' of the first motion to use it as the 'Initial Velocity' of the second motion which we can then use to get the second motion's displacement.
V1 = initial velocity
V2 = final velocity
First motion V2:
V2² = V1² + 2(a)(d)
V2² = (0m/s²) + 2(+2.6m/s²)(120m)
√V2² = √624 m/s
V2 = 24.9800 m/s
Now we find the time for the second motion of the journey.
Let V1 (initial velocity) be the final velocity of the first motion.
(V2 - V1)/a = t
(12 m/s² - 24.9800 m/s²) / -1.5 m/s² = t
t = 8.6533 s
Now we can calculate the second motion's displacement of our journey with the formula:
d = ((V1 + V2)/2) x t
d = ((24.9800 m/s² + 12 m/s²) / 2) x 8.6533 s
d = 160 m
Now add the first and second motion displacements:
160 m + 120 m = 280m
The total displacement of the motorcycle is 280m
Well, let's break it down, shall we?
First, the motorcycle starts from rest, so its initial velocity is 0 m/s. It then accelerates at +2.6 m/s^2 for a certain distance until it reaches a velocity of +12 m/s.
Using the equation:
v^2 = u^2 + 2as
where "v" is the final velocity, "u" is the initial velocity, "a" is the acceleration, and "s" is the displacement, we can solve for "s" using the given values.
12^2 = 0^2 + 2(2.6)(s)
144 = 5.2s
s ≈ 27.69 m
So, the motorcycle travels a distance of approximately 27.69 m during the acceleration phase.
Now, the motorcycle slows down with an acceleration of -1.5 m/s^2 until its velocity is +12 m/s. We can use the same equation to find the displacement during deceleration.
12^2 = 0^2 + 2(-1.5)(s)
144 = -3s
s ≈ -48 m
Wait, negative displacement? That's a bit odd. It seems like the motorcycle has traveled in the opposite direction during the deceleration phase.
To find the total displacement, we add the two displacements together:
27.69 m + (-48 m) = -20.31 m
So, the total displacement of the motorcycle is approximately -20.31 meters.
Hmmm... Seems like the motorcycle had a few detours along the way. Maybe it's practicing some clownish stunts?
To find the total displacement of the motorcycle, we need to find the distance traveled during the acceleration phase and the distance traveled during the deceleration phase.
First, let's find the time taken during the acceleration phase using the equation:
v = u + at
Here, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), a is the acceleration, and t is the time.
Since the final velocity is 12 m/s and the initial velocity is 0, we can rearrange the equation to solve for time:
t = (v - u) / a
Substituting the values:
t = (12 - 0) / 2.6
t = 4.62 s (rounded to two decimal places)
Next, let's calculate the distance traveled during the acceleration phase using the equation:
s = ut + (1/2)at^2
Here, s is the distance, u is the initial velocity (0), a is the acceleration (2.6), and t is the time (4.62):
s = 0 * 4.62 + (1/2) * 2.6 * (4.62)^2
s ≈ 26.29 m (rounded to two decimal places)
Now, let's find the time taken during the deceleration phase. Again, using the equation:
t = (v - u) / a
Here, the final velocity is 12 m/s, the initial velocity is 0, and the acceleration is -1.5 (negative since it is deceleration):
t = (12 - 0) / -1.5
t = -8 s
Since time cannot be negative, we know that this is not the correct answer. To find the correct time, we need to solve for time using the equation:
s = ut + (1/2)at^2
In this case, the distance traveled during deceleration is 120 m, the initial velocity is 12 m/s, and the acceleration is -1.5:
120 = 12 * t + (1/2) * -1.5 * t^2
This equation can be rearranged to a quadratic equation:
0 = 0.75t^2 - 12t + 120
Solving this quadratic equation, we get two possible values for time:
t = 9.33 s or t = 16 s
Since the motorcycle is slowing down, we can discard the larger time value (t = 16 s) as it would result in a negative final velocity. Therefore, the correct time is t = 9.33 s.
Now, let's calculate the distance traveled during the deceleration phase using the equation:
s = ut + (1/2)at^2
Here, s is the distance (120 m), u is the initial velocity (12 m/s), a is the acceleration (-1.5), and t is the time (9.33):
120 = 12 * 9.33 + (1/2) * -1.5 * (9.33)^2
120 ≈ 102.94 m (rounded to two decimal places)
Now, we can find the total displacement by adding the distances traveled during acceleration and deceleration:
Total displacement = Distance during acceleration + Distance during deceleration
Total displacement = 26.29 m + 102.94 m
Total displacement ≈ 129.23 m (rounded to two decimal places)
Therefore, the total displacement of the motorcycle is approximately 129.23 m.