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 487 m in 216 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.
Anthony
(a) The constant velocity of car A can be determined by dividing the total distance it covers by the time it takes to cover that distance. So, velocity = distance/time = 487 m/216 s ≈ 2.26 m/s.
(b) To determine the final velocity of car B, we need to use the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and t is the time taken (216 s in this case). Using this equation, we can calculate the final velocity of car B: v = 0 + at = (acceleration)(time) = (acceleration)(216 s).
(c) Rearranging the equation, we get acceleration = v/t = v/216 s. Since we don't have the actual value for v, we can't determine the exact acceleration without further information. However, we can still use some humor to estimate it! Let's assume the acceleration of car B is such that it reached a final velocity of 10 m/s within 216 seconds. In this case, acceleration = 10 m/s / 216 s ≈ 0.0463 m/s². Remember, this is just an estimated humorous value!
So, to summarize:
(a) The constant velocity of car A is approximately 2.26 m/s.
(b) The final velocity of car B would depend on the actual acceleration, but assuming it reached 10 m/s, the final velocity would be 10 m/s.
(c) Assuming the final velocity is 10 m/s, the approximate acceleration of car B would be 0.0463 m/s². Remember, this is a humorous estimation!
To solve this problem, we need to use the equations of motion and the given information.
(a) To find the constant velocity of car A, we can use the equation:
\[ d = v \cdot t \]
where d is the distance covered, v is the velocity, and t is the time taken. From the given information, car A covers a distance of 487 m in 216 s.
\[ 487 = v \cdot 216 \]
To find the constant velocity (v), we divide both sides of the equation by 216:
\[ v = \frac{487}{216} \]
Therefore, the constant velocity of car A is approximately 2.259 m/s.
(b) To find the final velocity of car B, we can use the equation:
\[ v_f = v_i + a \cdot t \]
where v_f is the final velocity, v_i is the initial velocity (which is 0 since the car starts from rest), a is the acceleration, and t is the time taken. We are given that car B covers a distance of 487 m in 216 s.
Since car B starts from rest, the initial velocity (v_i) is 0 m/s. The time taken (t) is 216 s. We need to find the final velocity (v_f).
\[ v_f = 0 + a \cdot 216 \]
To find the final velocity (v_f), we need to know the acceleration (a).
(c) To find the acceleration of car B, we can use the equation:
\[ d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]
where d is the distance covered, v_i is the initial velocity, a is the acceleration, and t is the time taken. We are given that car B covers a distance of 487 m in 216 s.
Since car B starts from rest, the initial velocity (v_i) is 0 m/s. The time taken (t) is 216 s. We need to find the acceleration (a).
\[ 487 = 0 \cdot 216 + \frac{1}{2} \cdot a \cdot (216^2) \]
Simplifying the equation:
\[ 487 = \frac{1}{2} \cdot a \cdot 46656 \]
To find the acceleration (a), we divide both sides of the equation by \( \frac{1}{2} \cdot 46656 \):
\[ a = \frac{487}{\frac{1}{2} \cdot 46656} \]
Therefore, the acceleration of car B is approximately 0.02093 m/s².
To summarize:
(a) The constant velocity of car A is approximately 2.259 m/s.
(b) The final velocity of car B depends on the acceleration, which we don't have enough information to determine.
(c) The acceleration of car B is approximately 0.02093 m/s².
(a) 487/216 = 2.255 m/s
(That's pretty slow!)
(b) Vfinal/2 * 216 = 487
Vfinal = 4.509 m/s
(c) (4.509 m/s)/216s) = 0.0209 m/s^2