How Can You Calculate Train Displacement in the First 3.15 Seconds of Motion?

[david12]

I try my best to solve this problem but couldn't ...here is the question

1.the train starting rest leaves a station with a constant acceleration.at the end of 5.34 s,it is moving at 15.549m/s. what is the train displacement in the first 3.15 s of motion?answer in units of meter.

I try in this way but couldn't get the answer.
I use velocity = displacement/time

15.549m/s= S/(5.34-3.15)

then i got 34.0523 m
after i got 34.05m i calculated the distance from 0s to 5.35 s which is 83.0316

finaly i subtract 83.0316 - 34.0523 = 48.97 m

but my answer is wrong so is there anyone can show me how i can do this question
thank you.

 

[Kurdt]

Are you familiar with the kinematic equations?

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

With the given information you need to first find the acceleration of the train and then the displacement after 3.15s.

 

[baba89]

I understand your frustration with this problem. It can be challenging to accurately solve for displacement when given limited information and a complex situation like a train starting with constant acceleration. I would suggest breaking down the problem into smaller steps and using the equations of motion to solve for displacement.

First, let's start with the given information. We know that the train has a constant acceleration, which means that its velocity is changing at a constant rate. We also know that after 5.34 seconds, the train is moving at a velocity of 15.549 m/s.

Next, let's use the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since the train was at rest), a is the acceleration, and t is the time. We can rearrange this equation to solve for the acceleration, which is a = (v-u)/t. Plugging in the values, we get a = (15.549 m/s - 0 m/s)/5.34 s = 2.911 m/s^2.

Now, we can use the equation s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. We are trying to find the displacement in the first 3.15 seconds, so we can plug in t = 3.15 s and solve for s. We get s = (0 m/s)(3.15 s) + 1/2(2.911 m/s^2)(3.15 s)^2 = 13.742 m.

Therefore, the train displacement in the first 3.15 seconds of motion is 13.742 meters. I hope this helps and shows you a different approach to solving this problem. Remember to always carefully analyze the given information and use the appropriate equations to solve for the unknown variables. Keep practicing and don't get discouraged, as problem-solving is a crucial skill in science.