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Particles Velocity, Displacement, and Total Distance Analysis

Created by @nathanedwards
 at October 31st 2023, 4:43:22 pm.
AP_Calculus_AB-Questions
#displacement
#time-interval
#velocity-function

Question:

The velocity function of a particle moving along a line is given by 𝑣(𝑡) = 4𝑡 − 3, where 𝑡 is measured in seconds and 𝑣(𝑡) is measured in meters per second.

a) Find the displacement of the particle over the time interval [0, 4].

b) Determine the time at which the particle is momentarily at rest during the interval [0, 4].

c) Find the total distance traveled by the particle over the time interval [0, 4].

Answer:

a) To find the displacement of the particle over the time interval [0, 4], we need to integrate the velocity function with respect to time:

Displacement=∫04(4t−3)dt \text{{Displacement}} = \int_{0}^{4} (4t - 3) dt

Using the power rule of integrals, we can integrate each term separately:

Displacement=[2t2−3t]04 \text{{Displacement}} = \left[2t^2 - 3t\right]_{0}^{4}

Evaluating the integral limits, we have:

Displacement=[2(4)2−3(4)]−[2(0)2−3(0)] \text{{Displacement}} = \left[2(4)^2 - 3(4)\right] - \left[2(0)^2 - 3(0)\right]
Displacement=32−0 \text{{Displacement}} = 32 - 0
Displacement=32 meters \text{{Displacement}} = 32 \, \text{{meters}}

Therefore, the displacement of the particle over the time interval [0, 4] is 32 meters.

b) To find the time at which the particle is momentarily at rest, we need to find the value of time where the velocity function is equal to 0:

4t−3=0 4t - 3 = 0

Solving for t, we have:

4t=3  ⟹  t=34 4t = 3 \implies t = \frac{3}{4}

Therefore, the particle is momentarily at rest at t=34t = \frac{3}{4} seconds.

c) To find the total distance traveled by the particle over the time interval [0, 4], we need to consider both positive and negative displacement. We can find the total distance by integrating the absolute value of the velocity function with respect to time:

Total Distance=∫04∣4t−3∣dt \text{{Total Distance}} = \int_{0}^{4} \left|4t - 3\right| dt

Since the velocity function 4t−34t - 3 is positive over the interval [0, 3/4) and negative over the interval (3/4, 4], we have:

Total Distance=[−(4t−3)]03/4+[(4t−3)]3/44 \text{{Total Distance}} = \left[-(4t - 3)\right]_{0}^{3/4} + \left[(4t - 3)\right]_{3/4}^{4}

Evaluating the integral limits, we get:

Total Distance=[−(4(3/4)−3)−(4(0)−3)]+[(4(4)−3)−(4(3/4)−3)] \text{{Total Distance}} = \left[-(4(3/4) - 3) - (4(0) - 3)\right] + \left[(4(4) - 3) - (4(3/4) - 3)\right]
Total Distance=(−0−(−3))+(13−(3−3)) \text{{Total Distance}} = (-0 - (-3)) + (13 - (3 - 3))
Total Distance=3+13 \text{{Total Distance}} = 3 + 13
Total Distance=16 meters \text{{Total Distance}} = 16 \, \text{{meters}}

Therefore, the total distance traveled by the particle over the time interval [0, 4] is 16 meters.

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