Confusing question to me

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This is because the acceleration is constant, so the average speed is just the midpoint of the initial and final speeds. So after 3 seconds, the average speed is (0 + 40)/2 = 20/2 = 10 m/s, and since it was 0 initially, the speed at 3 seconds is 10 m/s. So the distance traveled in the first 3 seconds is 10 m/s x 3 s = 30 m.
  • #1

dynamic998

A train starts from rest and accelerates uniformly in a straight line until, after 10s, it attains a speed of 40m/s. It then continues to accelerate at the same rate until it has traveled a total distance of 500m.

What is the distance travled by the train during the first 10s?
What distance did the train travel during the third second?

Ok this question is very confusing to me. I know that vi=0 and
vf=40 and distance= 500m. The thing i don't get is the first sentence. If u start from rest which is vi=0, how do u accelerate in a straight line to 10s. The thing i m guessing is if u go 4m/s from 0s to 10s. Then from 10s-22s, u travel at 40m/s. The first 10 seconds would equal to 20m, and the last 12 seconds would equal 480 m for a total of 500m. Does anyone know if i m doing this the right way because i feel that i got the beginning part messed up.
So for the first question i would think it is 20meters, and for the second question i think it is 12meters. Anyone can help?
 
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  • #2
dynamic998, there are some formulae for problems like this, but you can do without them here.

Remember, if speed goes up uniformly from 0 to 40, then the average speed must be 20. This, plus some common sense, is enough to solve the problem.
 
  • #3
Just some more info ...
The distance traveled in the third second is the distance traveled in the first three seconds minus the distance traveled in the first two seconds.
Try to take the question as two separate questions.
The first question says that train accelerated from rest to 40 m/s in 10 s.
The second question says that the train accelerated for 500 m, with a certain acceleration (that you would have calculated from the first question).
If you feel still stuck, tell us where exactly you are feeling stuck !
Ok this question is very confusing to me. I know that vi=0 and
vf=40 and distance= 500m. The thing i don't get is the first sentence. If u start from rest which is vi=0, how do u accelerate in a straight line to 10s. The thing i m guessing is if u go 4m/s from 0s to 10s. Then from 10s-22s, u travel at 40m/s. The first 10 seconds would equal to 20m, and the last 12 seconds would equal 480 m for a total of 500m. Does anyone know if i m doing this the right way because i feel that i got the beginning part messed up.
Frankly, i didn't understand what you mean by all of this !
 
  • #4
what i think

when i calculated this my answers for distance traveled in first 10s was 20m/s x 10s= 200meters. The distance traveled during first 3 seconds was 60m-40m= 20meters. Am i right?
 
  • #5
Some hints

It's a bit of a trick question. From what you are telling us, the fact it traveled in total 500m is completely useless.

Now, a number of equations that are helpful for these questions. The SUVAT equations, derived by calculus...
When v = final velocity, u = initial velocity, a = acceleration, t = time taken, s = displacement (change in distance in a certain direction)... And an object is traveling at CONSTANT acceleration...

v = u + at
v2 = u2 + 2*a*s
s = ut + 0.5 * a * t2
s = (u + v) * t /2
 
  • #6


Originally posted by dynamic998
distance traveled in first 10s was 20m/s x 10s= 200meters.
Correct.
The distance traveled during first 3 seconds was 60m-40m= 20meters.
Wrong. The speed is only 12 m/s after 3 seconds. So the distance is only 18 m.
 

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