Calculating Velocity and Displacement Using Acceleration Formula | Train Example

[MagMag]

I'm a little stuck on this:

Suppose that the acceleration of a train during the interval of time from t=2 s to t=4 s is a=2t m/s^2, and at t=2 s its velocity is 180km/h

Using formulas: Velocity v=ds/dt and acceleration a =dv/dt, determine:

(a) the train’s velocity at t=4 s,

and;

(b) the displacement (change in position) from time t=2 s to t= 4 s



I should really be able to do this but for some reason my mind has frozen so I was hoping for some help or some nudges in the right direction!

Thanks in advance!

 

[Stovebolt]

I'm a little stuck on this:

Suppose that the acceleration of a train during the interval of time from t=2 s to t=4 s is a=2t m/s^2, and at t=2 s its velocity is 180km/h

Using formulas: Velocity v=ds/dt and acceleration a =dv/dt, determine:

(a) the train’s velocity at t=4 s,

and;

(b) the displacement (change in position) from time t=2 s to t= 4 s



I should really be able to do this but for some reason my mind has frozen so I was hoping for some help or some nudges in the right direction!

Thanks in advance!

Given the way the question is asked, you should approach this as a calculus problem rather than a physics problem, so disregard the typical kinematics equations.

You are given a = 2t and a = dv/dt, so it follows that dv/dt = 2t. You are also given the interval of t = 2 to t = 4. Knowing the basics of integral calculus, how do you find the equation for V? (don't forget to add in the initial velocity)

Once you get the equation for V, and given that V = ds/dt, how do you find an equation for distance travelled?

 

[thomate1]

I can understand how the concept of velocity and displacement can be confusing at times. Let's break down this problem step by step to help you better understand.

First, we are given the acceleration of the train during the time interval of t=2 s to t=4 s, which is a=2t m/s^2. This means that the acceleration is changing over time, as it is dependent on the variable t.

Next, we are given the train's velocity at t=2 s, which is 180 km/h. However, we need to convert this to m/s to use in our calculations. We know that 1 km = 1000 m and 1 h = 3600 s. Therefore, 180 km/h = (180 * 1000) / (3600) = 50 m/s.

Now, let's use the formula v=ds/dt to find the train's velocity at t=4 s. We know that s is the displacement, and t is the time. Since we are given the acceleration and the initial velocity, we can use the formula v=u+at, where u is the initial velocity. Substituting in the values, we get:

v = (50 m/s) + (2 * 4 s) = 50 m/s + 8 m/s^2 = 58 m/s

Therefore, the train's velocity at t=4 s is 58 m/s.

Moving on to part (b), we need to find the displacement from t=2 s to t=4 s. We can use the formula s=ut+1/2at^2, where u is the initial velocity and a is the acceleration. Substituting in the values, we get:

s = (50 m/s * 2 s) + 1/2 * (2 * 4 s^2) = 100 m + 1/2 * 8 s^2 = 100 m + 16 m = 116 m

Therefore, the displacement from t=2 s to t=4 s is 116 m.

I hope this explanation helped you understand the problem better. Remember to always break down the problem into smaller steps and use the appropriate formulas to solve it. Keep practicing and you will become more comfortable with these concepts. Good luck!