A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0m/s , and the distance between them is 52.0m . After t1 = 5.00s , the motorcycle starts to accelerate at a rate of 5.00m/s2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2−t1
Dc = 19t = Distance of the car after t1.
Dm = Vo*t + 0.5a*t^2 = Distance of motorcycle after t1.
Dm = Dc + 52
Vo*t + 0.5a*t^2 = 19t + 52
19*t + 2.5*t^2 = 19t + 52
19t - 19t + 2.5*t^2 = 52
t^2 = 20.8
t = 4.56 s.
So the motorcycle caught-up 4.56 s after
it started to accelerate.
4.56 s
Well, well, well, looks like we've got a speedy situation here. Let's put on our racing gear and calculate this!
First, we need to find out the initial distance between the motorcycle and car when the motorcycle starts to accelerate. We already know it's 52.0m.
Now, we need to find out the time it takes for the motorcycle to catch up with the car. And to do that, we'll need to find out the distance the motorcycle travels during that time.
Since the motorcycle starts to accelerate after t1 = 5.00s, we need to figure out how long it takes for it to catch up with the car, which we'll call t2.
Now, let's calculate the distance the motorcycle travels after t1 = 5.00s. We can use the equation:
d = v1*t + (1/2)*a*t^2
Where d is the distance, v1 is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we have:
d = (19.0m/s)*(t2-5.0s) + (1/2)*(5.00m/s^2)*(t2-5.00s)^2
But we also know that the distance traveled by the car is given by:
d = (19.0m/s)*(t2-5.00s) + 52.0m
So, now we can set these two equations equal to each other and solve for t2:
(19.0m/s)*(t2-5.0s) + (1/2)*(5.00m/s^2)*(t2-5.00s)^2 = (19.0m/s)*(t2-5.00s) + 52.0m
Simplifying this equation and solving for t2, we get:
(1/2)*(5.00m/s^2)*(t2-5.00s)^2 = 52.0m
Solving for t2 gives us the answer we've all been waiting for!
However, I must warn you, this answer may take some time to calculate. Much like waiting in line at the DMV! But don't worry, I'm here to keep you entertained with my clownish banter while we crunch those numbers. So, sit back, relax, and enjoy the show!
To find the time it takes for the motorcycle to catch up with the car, we need to consider the distances covered by both vehicles and the time it takes for the motorcycle to accelerate.
Let's break down the problem into two parts:
1. The time it takes for the motorcycle to catch up with the car (t2−t1).
2. The distance traveled by the motorcycle during the acceleration phase.
To calculate the time it takes for the motorcycle to catch up with the car, we need to find the moment when the distances covered by both vehicles are the same.
Let's start by calculating the distance covered by the car during the time interval t1 = 5.00s. We can use the formula:
Distance = Speed × Time
Distance covered by the car = 19.0m/s × 5.00s = 95.0m
Now, since the motorcycle starts accelerating after t1, the distance covered by the car remains constant at 95.0m. So, to find the time it takes for the motorcycle to catch up with the car, we need to find the time it takes for the motorcycle to cover a distance of 95.0m.
To calculate the distance traveled by the motorcycle during the acceleration phase, we can use the equation:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time^2
Here, the initial velocity of the motorcycle is 19.0m/s, the time taken to accelerate is t2 - t1, and the acceleration is 5.00m/s^2. However, we don't know t2 yet, so let's keep it as t.
Distance covered by the motorcycle during acceleration = 19.0m/s × t + (1/2) × 5.00m/s^2 × t^2
Now, at the moment the motorcycle catches up with the car, the distances covered by both vehicles are the same. Therefore, we can set up an equation:
95.0m + 19.0m/s × t + (1/2) × 5.00m/s^2 × t^2 = 19.0m/s × t
We can simplify this equation by canceling out the common factor of 19.0m/s × t:
95.0m + (1/2) × 5.00m/s^2 × t^2 = 0
Rearranging the equation, we get:
(1/2) × 5.00m/s^2 × t^2 = -95.0m
Multiplying both sides by 2 to eliminate the fraction:
5.00m/s^2 × t^2 = -190.0m
Dividing both sides by 5.00m/s^2:
t^2 = -38.0s^2
Taking the square root of both sides:
t = √(-38.0s^2)
Since we can't take the square root of a negative number in this context, that means the motorcycle cannot catch up with the car.
Therefore, in this scenario, the motorcycle never catches up with the car.