Two motorcycles are at rest and separated by 24.5 ft. They start at
the same time in the same direction, the one in the back having an
acceleration of 3 ft. per sec2, the one in the front going slower at an
acceleration of 2 ft. per sec2.
(a) How long does it take for the faster cycle to overtake the slower?
(b)How far does the faster machine go before it catches up?
(c) How fast is each cycle going at this time?
(a) 1/2 * 3t^2 = 24.5 + 1/2 * 2t^2
(b) the above expressions are the distances covered.
(c) v = at for each cycle.
Oh, motorcycles racing! Let's calculate the clowny way!
(a) How long does it take for the faster cycle to overtake the slower one?
Well, first we need to find out when they meet. We can use some math tricks or just rely on our good friend, common sense. The faster motorcycle is gaining 1 foot per sec every second on the slower one (3 ft. per sec2 – 2 ft. per sec2). So, it will take 24.5 seconds for the faster motorcycle to catch up to the slower one. Why? Because 24.5 feet divided by 1 foot per second gives us... 24.5 seconds! Ta-da!
(b) How far does the faster machine go before it catches up?
The faster motorcycle travels for 24.5 seconds before reaching the slower one. Now, in this time, the motorcycle has been accelerating at 3 ft. per sec2. We can calculate the distance traveled using the formula: distance = 0.5 * acceleration * time^2. Plugging in the values, we get: distance = 0.5 * 3 ft. per sec2 * (24.5 sec)^2. So, the faster motorcycle goes approximately 903.94 feet before it catches up. That's quite a ride!
(c) How fast is each motorcycle going at this time?
At the moment when the faster motorcycle catches up, it has been accelerating at 3 ft. per sec2 for 24.5 seconds. So, its speed would be 3 ft. per sec2 * 24.5 seconds = 73.5 ft. per sec. As for the slower motorcycle, it has been accelerating at 2 ft. per sec2 for the same time. So its speed would be 2 ft. per sec2 * 24.5 seconds = 49 ft. per sec.
Remember, life is like a motorcycle race – full of acceleration, distances, and a sprinkle of physics!
To solve this problem, we can use the equations of motion to analyze the motion of the two motorcycles.
(a) To find how long it takes for the faster motorcycle to overtake the slower one, we can use the following equation:
s = ut + (1/2)at^2
Where:
s = separation between the motorcycles (24.5 ft)
u = initial velocity (0 ft/sec, as both motorcycles are at rest)
a = acceleration
For the slower motorcycle:
From the equation above, we have:
24.5 = (1/2) * 2 * t^2
49 = 2 * t^2
t^2 = 24.5
Taking the square root of both sides, we get:
t = √24.5
t ≈ 4.95 seconds
Therefore, the faster motorcycle takes approximately 4.95 seconds to overtake the slower one.
(b) To find how far the faster motorcycle goes before it catches up, we can use the equation:
s = ut + (1/2)at^2
For the faster motorcycle:
s = distance traveled
u = initial velocity (0 ft/sec, as both motorcycles are at rest)
a = acceleration (3 ft/sec^2, faster motorcycle)
From the equation above, we have:
s = 0 * t + (1/2) * 3 * t^2
s = (3/2) * t^2
Substituting the previously calculated value of t (t ≈ 4.95), we get:
s = (3/2) * (4.95)^2
s ≈ 36.53 ft
Therefore, the faster motorcycle travels approximately 36.53 ft before it catches up.
(c) To determine the speed of each motorcycle when the faster one catches up, we can use the equation:
v = u + at
For the slower motorcycle:
u = initial velocity (0 ft/sec, as both motorcycles are at rest)
a = acceleration (2 ft/sec^2, slower motorcycle)
t = time taken to catch up (approximately 4.95 seconds)
From the equation above, we have:
v = 0 + 2 * 4.95
v ≈ 9.9 ft/sec
For the faster motorcycle:
u = initial velocity (0 ft/sec, as both motorcycles are at rest)
a = acceleration (3 ft/sec^2, faster motorcycle)
t = time taken to catch up (approximately 4.95 seconds)
From the equation above, we have:
v = 0 + 3 * 4.95
v ≈ 14.85 ft/sec
Therefore, at the time the faster motorcycle catches up, the slower motorcycle is traveling at approximately 9.9 ft/sec, and the faster motorcycle is traveling at approximately 14.85 ft/sec.
To answer these questions, we need to make use of the equations of motion. The equations of motion can be derived from the basic kinematic equations, which relate the displacement, velocity, time, and acceleration of an object.
(a) To find how long it takes for the faster motorcycle to overtake the slower one, we need to determine when their displacements become equal. Let's assume that the faster motorcycle overtakes the slower motorcycle after time 't'.
The displacement of an object can be calculated using the equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
For the slower motorcycle:
displacement1 = 0 * t + (1/2) * 2 * t^2
For the faster motorcycle:
displacement2 = 24.5 ft + 0 * t + (1/2) * 3 * t^2
Setting these two displacements equal to each other:
0 * t + (1/2) * 2 * t^2 = 24.5 ft + 0 * t + (1/2) * 3 * t^2
Simplifying the equation:
t^2 = 24.5 ft / (1.5 ft/sec^2 - 1 ft/sec^2)
t^2 = 24.5 ft / 0.5 ft/sec^2
Now we can solve for 't':
t^2 = 49 sec^2
t = sqrt(49 sec^2)
t = 7 sec
Therefore, it takes 7 seconds for the faster motorcycle to overtake the slower one.
(b) To find how far the faster motorcycle goes before catching up, we need to calculate the displacement of the faster motorcycle at time 't'.
Using the equation of motion for the faster motorcycle:
displacement2 = 24.5 ft + 0 * 7 sec + (1/2) * 3 * (7 sec)^2
Simplifying the equation:
displacement2 = 24.5 ft + 0 ft + (1/2) * 3 * 49 ft
displacement2 = 24.5 ft + 0 ft + 73.5 ft
displacement2 = 98 ft
Therefore, the faster motorcycle goes 98 ft before catching up.
(c) To find the speed of each motorcycle at the time of catching up, we need to calculate their velocities using the equation:
velocity = initial velocity + acceleration * time
For the slower motorcycle:
velocity1 = 0 ft/sec + 2 ft/sec^2 * 7 sec
velocity1 = 14 ft/sec
For the faster motorcycle:
velocity2 = 0 ft/sec + 3 ft/sec^2 * 7 sec
velocity2 = 21 ft/sec
Therefore, at the time of catching up, the slower motorcycle is traveling at 14 ft/sec, and the faster motorcycle is traveling at 21 ft/sec.