Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest and maintains a constant acceleration. Both cars cover a distance of 515 m in 229 s. Assume that they are moving in the +x direction. Determine (a) the constant velocity of car A, (b) the final velocity of car B, and (c) the acceleration of car B.
a) 515/229
b) average speed = (0 + final speed)/2 = answer to part (a)
in other words
answer to b = twice answer to a
c) a = change in velocity/change in time
= answer to part b /229 seconds
To solve this problem, let's first define the variables:
Let V_A be the constant velocity of car A.
Let V_Bf be the final velocity of car B.
Let a_B be the constant acceleration of car B.
Let d be the distance covered by both cars, which is 515 m.
Let t be the time taken to cover the distance, which is 229 s.
Now, let's solve for each part of the problem:
(a) The constant velocity of car A (V_A):
Since car A covers the distance at a constant velocity, we can use the formula:
Velocity = Distance / Time
So, V_A = d / t = 515 m / 229 s ≈ 2.25 m/s.
Therefore, the constant velocity of car A is approximately 2.25 m/s.
(b) The final velocity of car B (V_Bf):
We know that the initial velocity of car B (V_Bi) is 0 m/s because it starts from rest.
We also know that the time taken (t) is 229 s.
Using the formula for final velocity with constant acceleration:
V_Bf = V_Bi + (a_B * t)
Since V_Bi = 0 m/s, we have:
V_Bf = a_B * t
Substituting the given values: 2.25 m/s = a_B * 229 s
Therefore, the final velocity of car B is 2.25 m/s.
(c) The acceleration of car B (a_B):
Using the same formula as before, we can rearrange it:
a_B = (V_Bf - V_Bi) / t
Substituting the given values: a_B = (2.25 m/s - 0 m/s) / 229 s
Therefore, the acceleration of car B is approximately 0.0098 m/s^2.
To summarize:
(a) The constant velocity of car A is approximately 2.25 m/s.
(b) The final velocity of car B is 2.25 m/s.
(c) The acceleration of car B is approximately 0.0098 m/s^2.
To solve this problem, we can use the equations of motion for uniformly accelerated motion.
(a) To find the constant velocity of car A, we can use the equation:
Distance = Velocity × Time
Since car A covers a distance of 515 m in 229 s, we can rearrange the equation to solve for velocity:
Velocity = Distance / Time
Plugging in the values, we get:
Velocity of car A = 515 m / 229 s
Velocity of car A ≈ 2.25 m/s
Therefore, the constant velocity of car A is approximately 2.25 m/s.
(b) To find the final velocity of car B, we can use the equation of uniformly accelerated motion:
Final Velocity = Initial Velocity + (Acceleration × Time)
We know that the initial velocity of car B is 0 m/s (as it starts from rest), and the time taken is 229 s. So, the equation becomes:
Final Velocity = 0 m/s + (Acceleration × 229 s)
We still need to find the acceleration to solve this equation.
(c) To find the acceleration of car B, we can use another equation of uniformly accelerated motion:
Distance = Initial Velocity × Time + (0.5 × Acceleration × Time^2)
We know that the distance covered is 515 m, the initial velocity is 0 m/s, and the time taken is 229 s. So, the equation becomes:
515 m = 0 m/s × 229 s + (0.5 × Acceleration × (229 s)^2)
Simplifying the equation, we get:
515 m = 0.5 × Acceleration × (229 s)^2
Now we can solve for the acceleration:
Acceleration = (2 × Distance) / (Time^2)
Plugging in the values, we get:
Acceleration = (2 × 515 m) / (229 s)^2
Acceleration ≈ 0.047 m/s^2
Therefore, the acceleration of car B is approximately 0.047 m/s^2.
Now that we have determined the constant velocity of car A (approximately 2.25 m/s), the final velocity of car B (which we can find by substituting the known values into the equation Final Velocity = 0 m/s + (Acceleration × Time)), and the acceleration of car B (approximately 0.047 m/s^2), we have successfully solved the problem.