RaySix

Post

Car Motion: Acceleration, Displacement, and Distance

Created by @nathanedwards

at November 1st 2023, 2:12:27 pm.

AP_Physics_1-Questions

#acceleration

#displacement

#velocity-time-graph

Question:

A car starts from rest and accelerates uniformly in a straight line. The velocity-time graph for the car's motion is given below:

Velocity-Time Graph

a) Determine the acceleration of the car during the time interval (0 s to 5 s).

b) Determine the displacement of the car during the time interval (0 s to 5 s).

c) Determine the distance covered by the car during the time interval (0 s to 5 s).

Answer:

a) To find the acceleration of the car during the time interval (0 s to 5 s), we need to calculate the slope of the velocity-time graph.

The slope of a velocity-time graph represents the rate of change of velocity, which is the acceleration.

Dividing the change in velocity by the change in time, we can determine the acceleration:

\text{Acceleration} = \frac{{\text{Change in Velocity}}}{{\text{Change in Time}}}

Looking at the graph, from time t = 0 s to t = 5 s, the change in velocity is 20 m/s - 0 m/s = 20 m/s.

The change in time is 5 s - 0 s = 5 s.

Substituting these values into the formula:

\text{Acceleration} = \frac{{20 \, \text{m/s} - 0 \, \text{m/s}}}{{5 \, \text{s} - 0 \, \text{s}}} = \frac{{20 \, \text{m/s}}}{{5 \, \text{s}}} = 4 \, \text{m/s}^2

Therefore, the acceleration of the car during the time interval (0 s to 5 s) is 4 m/s^2.

b) To determine the displacement of the car during the time interval (0 s to 5 s), we need to find the area under the velocity-time graph.

Looking at the graph, we can see that the graph represents a triangle. The area of a triangle is given by the formula:

\text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}

The base of the triangle is the time interval (5 s - 0 s) = 5 s.

The height of the triangle is the maximum velocity reached (20 m/s).

Plugging these values into the formula:

\text{Displacement} = \frac{1}{2} \times 5 \, \text{s} \times 20 \, \text{m/s} = 50 \, \text{m}

Therefore, the displacement of the car during the time interval (0 s to 5 s) is 50 m.

c) To determine the distance covered by the car during the time interval (0 s to 5 s), we need to find the total area between the velocity-time graph and the time axis.

Looking at the graph, we can see that the total area is the sum of the area of the triangle and the area of the rectangle.

The area of the triangle was calculated in part (b) as 50 m.

The area of the rectangle can be found using the formula:

\text{Area of Rectangle} = \text{Base} \times \text{Height}

The base of the rectangle is the time interval (5 s - 0 s) = 5 s.

The height of the rectangle is the constant velocity reached (20 m/s).

Plugging these values into the formula:

\text{Area of Rectangle} = 5 \, \text{s} \times 20 \, \text{m/s} = 100 \, \text{m}

Therefore, the total distance covered by the car during the time interval (0 s to 5 s) is:

\text{Distance} = \text{Area of Triangle} + \text{Area of Rectangle} = 50 \, \text{m} + 100 \, \text{m} = 150 \, \text{m}

Therefore, the distance covered by the car during the time interval (0 s to 5 s) is 150 m.