RaySix

Post

Cars Total Displacement with Uniform Acceleration

Created by @nathanedwards

at November 3rd 2023, 9:24:41 pm.

AP_Physics_1-Questions

#displacement

#uniform-acceleration

#constant-velocity

Question:

A car accelerates uniformly from rest to a speed of 30 m/s in a time of 10 seconds. The car then maintains a constant velocity of 30 m/s for 20 seconds. Finally, the car decelerates uniformly and comes to a stop in 5 seconds. Determine the total displacement of the car during this time interval.

Solution:

To solve this problem, we need to find the displacement during each phase of motion and then add them together.

First, we will find the displacement during the first phase when the car is accelerating uniformly. We are given that the initial velocity, $v_{0}$ , is 0 m/s, the final velocity, $v$ , is 30 m/s, and the time, $t$ , is 10 seconds. The displacement during this phase can be found using the formula:

d = \frac{1}{2}(v_{0} + v)t

Substituting the given values into the equation, we find:

d_{1} = \frac{1}{2}(0 + 30)(10) = 150 \, \text{m}

Next, we will find the displacement during the second phase when the car maintains a constant velocity of 30 m/s. The velocity during this phase is constant, so the displacement is given by:

d_{2} = v \cdot t

Substituting the given values, we get:

d_{2} = 30 \, \text{m/s} \cdot 20 \, \text{s} = 600 \, \text{m}

Finally, we will find the displacement during the third phase when the car decelerates uniformly and comes to a stop. We are given that the initial velocity is 30 m/s, the final velocity is 0 m/s, and the time is 5 seconds. Using the same formula as in the first phase, we can calculate the displacement:

d_{3} = \frac{1}{2}(30 + 0) \cdot 5 = 75 \, \text{m}

Now we can find the total displacement by adding the displacements from each phase:

d_{\text{total}} = d_{1} + d_{2} + d_{3} = 150 \, \text{m} + 600 \, \text{m} + 75 \, \text{m} = 825 \, \text{m}

Therefore, the total displacement of the car during this time interval is 825 meters.