# 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?

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

Get this polynomial to fit the known data - i.e velocity at particular time invervals.

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,

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Fermat
Homework Helper
How do we know that the accln is constant ?

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,

what about the instantaneous aceleration at 1.5?

What about it, he needs to find instantaneous velocity at 1.5 seconds, not acceleration.

Regards,

Fermat
Homework Helper
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
==========

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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,