Two cars both cover a straight distance, d = 268 m, in time t = 29 s. Car A moves at a constant velocity (vA). Car B moves at a constant acceleration (aB), starting from an initial velocity of v0B = 12 m/s. Assume both cars are moving in the positive x-direction.
What is the final velocity of Car B?
What is the acceleration of Car B?
s = vt + 1/2 at^2
12*29 + 1/2 a*29^2 = 268
a = -0.19 m/s^2
v = 12 + at = 12 - 0.19*29 = 6.49 m/s
Why do I suspect a typo?
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Well, let's find the final velocity of Car B first. Since Car B is moving with constant acceleration, we can use the equation vB = v0B + aB*t, where vB is the final velocity of Car B. We are given that the initial velocity v0B is 12 m/s and the time t is 29 seconds. Plugging in these values, we get vB = 12 m/s + aB*29 s. Unfortunately, we don't know the acceleration yet, so we can't determine the final velocity of Car B. Sorry, but it looks like Car B's final velocity will remain a mystery.
Now, let's move on to the acceleration of Car B. We can use the equation d = v0B*t + (1/2)*aB*t^2, where d is the distance covered by Car B. We are given that the distance d is 268 meters, the initial velocity v0B is 12 m/s, and the time t is 29 seconds. Plugging in these values, we get 268 m = (12 m/s)*29 s + (1/2)*aB*(29 s)^2. Solving for aB, we find aB = (268 m - 12 m/s * 29 s) / [(1/2)*(29 s)^2], which simplifies to aB = (268 m - 348 m) / (841 s^2), or aB = -80 m / 841 s^2.
So, the acceleration of Car B is approximately -0.095 m/s^2. Remember, though, laughter is always positive, even when acceleration is negative!
To find the final velocity of Car B, we can use the equation of motion:
vB = v0B + aB * t
where vB is the final velocity of Car B, v0B is the initial velocity of Car B, aB is the acceleration of Car B, and t is the time taken.
Given:
v0B = 12 m/s
t = 29 s
Substituting these values into the equation, we have:
vB = 12 m/s + aB * 29 s
To find the acceleration of Car B, we can use the formula:
d = v0B * t + 0.5 * aB * t^2
where d is the distance covered, v0B is the initial velocity, t is the time taken, and aB is the acceleration.
Given:
d = 268 m
v0B = 12 m/s
t = 29 s
Substituting these values into the equation, we have:
268 m = 12 m/s * 29 s + 0.5 * aB * (29 s)^2
Simplifying this equation, we get:
268 m = 348 m/s + 0.5 * aB * 841 s^2
Rearranging the equation to isolate the acceleration term, we have:
0.5 * aB * 841 s^2 = 268 m - 348 m/s
0.5 * aB * 841 s^2 = -80 m
Dividing both sides of the equation by 0.5 * 841 s^2, we have:
aB = (-80 m) / (0.5 * 841 s^2)
aB = -0.190 m/s^2
So the final velocity of Car B is vB = 12 m/s + aB * 29 s = 12 m/s + (-0.190 m/s^2) * 29 s = 12 m/s - 5.51 m/s = 6.49 m/s, and the acceleration of Car B is aB = -0.190 m/s^2.