Distance Covered While Acc, and Dec

  • Thread starter granitkr
  • Start date
In summary: Also, your formua has 't*s' in it which is not correct.In summary, the subway train starts from rest and accelerates at a rate of 1.20 m/s^2 for 20.0 seconds, covers a distance of 240 meters. It then runs at a constant speed for 50.0 seconds before decelerating at a rate of -3.00 m/s^2 until it stops at the next station. The total distance covered is 240 meters while accelerating, 600 meters while running at constant speed, and 450 meters while decelerating, for a total distance of 1,290 meters. The formula used to calculate the distance covered is D = 0.5
  • #1
granitkr
3
0
A subway train starts from rest at a station and accelerates at a rate of 1,20 m/s^2
for 20,0s. It runs at constant speed for 50,0 s and slows down at -3,00 m/s^2
until it stops at the next station. Find the total distance covered.

I just need to know the formulas, because i studied this a long time ago, and forgot them!
 
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  • #2
Ive been using this formula but somehow it dose not check out i THINK. (acc) D= 0.5 * 1.20 * 20^2
 
  • #3
Break the problem into several parts.

1. Use your formula above to determine how far it went when accelerating.
2. Use formula V=a*t to determine the speed attained in 1 above.
3. Use formula X = V*t to determine how far it went in the 50 seconds.
4. Use formua X = V*t + .5*a*t^2 to determine how far it went during the slowing down portion of the problem. Note that 'a' is -3.0 in this equation.
5. Add all the distances up.
 
  • #4
I still need the formula to calculate the distance traveled while acc. to 24m/s^2. Or is mine right D= 1/2 * a * t^2
 
  • #5
granitkr said:
I still need the formula to calculate the distance traveled while acc. to 24m/s^2. Or is mine right D= 1/2 * a * t^2

Recall my first statement:
1. Use your formula above to determine how far it went when accelerating.

You have the proper formula to determine distance for the acceleration portion.

Secondly your statement above does not make sense. The last part of it says "while acc. to 24m/s^2". You are implying velocity so the units are wrong.

V = a*t = (m/s^2)*s = m/s
 

Related to Distance Covered While Acc, and Dec

1. What is "Distance Covered While Acc, and Dec"?

Distance Covered While Acc, and Dec is a measure of the total distance traveled while accelerating and decelerating. It takes into account the speed of an object, the time it takes to accelerate or decelerate, and the distance traveled during those periods.

2. How is "Distance Covered While Acc, and Dec" calculated?

The formula for calculating Distance Covered While Acc, and Dec is: D = (Vf² - Vi²) / 2a, where D is the distance, Vf is the final velocity, Vi is the initial velocity, and a is the acceleration.

3. Why is "Distance Covered While Acc, and Dec" important in physics?

Distance Covered While Acc, and Dec is important in physics because it helps us understand the motion of objects and how they accelerate and decelerate. It also allows us to calculate the distance an object has traveled during these periods, which is essential in many real-world applications.

4. Can "Distance Covered While Acc, and Dec" be negative?

Yes, "Distance Covered While Acc, and Dec" can be negative if the initial and final velocities have opposite signs. This means that the object is decelerating during the given time period, resulting in a negative distance.

5. How does air resistance affect "Distance Covered While Acc, and Dec"?

Air resistance can affect "Distance Covered While Acc, and Dec" by slowing down the object's acceleration and deceleration. This is because air resistance creates a force that opposes the motion of the object, making it more difficult for the object to accelerate or decelerate. As a result, the distance covered while accelerating and decelerating may be shorter than expected.

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