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An acceleration problem

This problem request the instanteous velocity at a given point which I have not yet discovered how to do, the problem is as follows...

The engine of a model rocket accelerates the rocket vertically upward for 2.0 s as follows: at t = 0, the rocket's speed is zero; at t = 1.0 s, its speed is 4.0 m/s; and at t = 2.0 s, its speed is 20 m/s. Plot a velocity vs. time graph for this motion.

I have found that the average acceleration during the 2.0 s interval is 10 m/s2

I am trying to find the instanteous velocity at 1.5. I know that i must find the slope of the line tangent to the point 1.5 on the graph but am having problms doing so. Any tips?
 

Answers and Replies

Fermat
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try representing the accln by a polynomial, in time.

Get this polynomial to fit the known data - i.e velocity at particular time invervals.
 
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Fermat said:
try representing the accln by a polynomial, in time.

Get this polynomial to fit the known data - i.e velocity at particular time invervals.
Since the velocity rate of change varies at the two intervals, and we are dealing with constant acceleration, then I assume you mean try to find two different lines for the two intervals and not one polynomial.

You need to find the slope of the velocity time surve between 0 and 1 second, the do the same for the interval of 1 to 2 seconds. Once you have this, read the graph and enjoy.


Regards,

Nenad
 
Last edited:
Fermat
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How do we know that the accln is constant ?
 
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Im assuming it is, but you might be right, since it is a rocket, acceleration might vary, but I doubt it since no mass is lost. Its not a momentum problem where they give you the mass flow rate out, I think its a simple kinematics problem, thats why I assumed constant acceleration.

Regards,

Nenad
 
what about the instantaneous aceleration at 1.5?
 
698
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What about it, he needs to find instantaneous velocity at 1.5 seconds, not acceleration.

Regards,

Nenad
 
Fermat
Homework Helper
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If it's constant acceleration, there are a couple of ways.

You plotted the velocity-time graph - no ?
If so, then just read off the graph at t = 1.5s. You should get v = 12 m/s

Alternatively, you mentioned getting the slope of the graph between 1 and 2 secs.
let m be the slope. Then m = (v2 - v1)/(t2 - t1), where v1 is the velocity at t1 and v2 is the velocity at t2, and t1 = 1 sec and t2 = 2 sec.
This should give you m - the acceleration during the 2nd second of movement. Now simply use v = u + at over this movement.

If the accelerations aren't constant, and it is a rocket after all, then you can approximate its motion over the 2-second interval by a polynomial.

Let the velocity, v = at² + bt.
we have two data points, (1,4) and (2,20), where (x,y) == (time,velocity)
substituting for these data points in the eqn for velocity,

4 = a + b
20 = 4a + 2b

giving,

a = 6, b = -2

.: v = 6t² - 2t
v = 2t(3t - 1)
===========

at t = 1.5
v = 3(4.5 - 1)
v = 10.5 m/s
==========
 
Last edited:
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Fermat said:
If it's constant velocity, there are a couple of ways.
Im assuming you meant to write "constant acceleration". The rest looks correct.

Regards,

Nenad
 
Fermat
Homework Helper
872
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Nenad said:
Im assuming you meant to write "constant acceleration". The rest looks correct.

Regards,

Nenad
Thanks for the note.

Edited my post
 

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