A car and train move together along parallel paths at 25 m/s [W]. The car then undergoes a uniform acceleration of 2.5m/s^2 [E] because of a red light and comes to rest. It remains at rest for 45s, then accelerates back to a velocity of 25m/s [W] at a rate of 2.5m/s^2 [W]. How far behind the train is the car when it reaches the velocity of 25m/s [W], assuming that the train's velocity has remained constant?
Why did the car and train become such good friends? Because they had parallel paths! Now, let's calculate how far behind the train the car is when it reaches a velocity of 25m/s [W].
First, let's determine how long it takes for the car to come to rest. Using the acceleration and initial velocity, we can use the formula:
v = u + at
where:
v = final velocity (0 m/s, since the car comes to rest)
u = initial velocity (25 m/s [W])
a = acceleration (-2.5 m/s^2 [E])
t = time
Rearranging the formula, we have:
t = (v - u) / a
t = (0 - 25) / -2.5
t = 10 seconds
So, it takes 10 seconds for the car to come to a stop. After that, it remains at rest for 45 seconds. So, the total time from the initial velocity of 25 m/s [W] to when it starts accelerating back to the same velocity is 10 + 45 = 55 seconds.
Now, let's find out the distance the car travels during this time. We can use the formula:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity (0 m/s, since the car starts from rest)
a = acceleration (2.5 m/s^2 [W])
t = time (55 seconds)
s = 0 * 55 + (1/2) * 2.5 * (55^2)
s = 1/2 * 2.5 * 3025
s = 2.5 * 1512.5
s = 3781.25 meters
So, the car travels a distance of 3781.25 meters behind the train when it reaches a velocity of 25 m/s [W]. Keep your eyes on the road and don't get distracted by red lights!
To find how far behind the train the car is when it reaches a velocity of 25 m/s [W], we need to break down the car's motion into different stages and calculate the distance traveled in each stage.
Let's analyze the car's motion:
1. Stage 1: Car moves with constant velocity at 25 m/s [W] along with the train.
- The time taken in this stage is not given, so we'll consider it as 't'.
Distance covered in Stage 1 = Velocity × Time = 25 m/s × t = 25t meters [E] from the starting point.
2. Stage 2: Car undergoes a uniform deceleration and comes to rest.
- The acceleration of the car is given as -2.5 m/s^2 [E].
- The initial velocity is 25 m/s [W].
- We need to find the time taken in this stage.
Using the equation of motion: v = u + at (where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time)
0 = 25 m/s - 2.5 m/s^2 × t
Solving this equation, we find:
t = 10 seconds
Distance covered in Stage 2 can be calculated using the equation of motion: s = ut + (1/2)at^2
s = 25 m/s × 10 s + (1/2) × (-2.5 m/s^2) × (10 s)^2 = 250 meters [E] from the starting point.
3. Stage 3: Car remains at rest for 45 seconds.
- The car is stationary, so it doesn't cover any distance in this stage.
4. Stage 4: Car accelerates back to a velocity of 25 m/s [W].
- The acceleration of the car is given as 2.5 m/s^2 [W].
- We need to find the time taken in this stage.
Using the same equation of motion: v = u + at
25 m/s = 0 + 2.5 m/s^2 × t
Solving this equation, we find:
t = 10 seconds
Distance covered in Stage 4 can be calculated using the equation of motion: s = ut + (1/2)at^2
s = 0 + (1/2) × 2.5 m/s^2 × (10 s)^2 = 125 meters [W] from the starting point.
Now, we can find the total distance covered by the car:
Total distance = Distance in Stage 1 + Distance in Stage 2 + Distance in Stage 3 + Distance in Stage 4
= 25t + 250 + 0 + 125
= 25t + 375 meters
To find the distance behind the train when the car reaches a velocity of 25 m/s, we need to substitute the value of 't' into the equation:
Distance behind = 25t + 375
Since the time taken in Stage 1 is not given, we cannot determine the exact distance behind the train without that information.