Average Velocity: particle moved according to s(t) = t 2 + cost (t) / t4 + 1

[sirhc]

Hey,

Just need a little help with this, (it just seems wrong)

"A particle is moving along a straight line such that its position function s(t) at time t seconds is given by

s(t) = t ² + cost (t) / t⁴ + 1

Determine the average velocity of the particle over the first 2 seconds"

My attempt so far..... Average velocity = s(2²+cos(2) / (2)⁴+1) - s(0²+cos(0) / (0)⁴+1) / 2-0 = - 13/34 (-0.3823) | using the average velocity formula (s(t₂) - s(t₁) / t₂ - t₁

Thanks

 

[HallsofIvy]

The way you have written this is very confusing. First, do you mean that s(t) = (t^2 + cos (t)) / (t^4 + 1)?

Second, the "average velocity of the particle over the first 2 seconds" is (s(2)- s(0))/2, NOT "s(0^+ cos(0))" etc. Of course that is [(2^2+ cos(2))/(2^4+ 1)- (0^2+ cos(0))/(0^4+ 1)]/2= [(4+ cos(2))/5- (0+ cos(0)/1)]/2= [4/5+ cos(2)/5- 1]/2= (cos(2)- 1/5)/2.

 

[sirhc]

Sorry about that;

(1) A particle is moving along a straight line such that its position function s(t) at time t seconds is given by

s(t) = (t^2 + cos (t)) / (t^4 + 1)

a)Determine the average velocity of the particle over the first 2 seconds

(s(2²+cos(2))) / ((2)⁴+1) - (s(0²+cos(0))) / ((0)⁴+1) / (2-0)

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Can the average velocity be a negative...?

 

[tkhunny]

Sure! Once you decide which direction is positive, then the other direction must be negative.

Demonstrate your intermediate results. Be confident in your solution. What do you get? Does it make sense?

 

[sirhc]

- 13/34 (-0.3823)


Does that mean you just change it to positive. (e.g. = 0.3823 metres after two seconds)

 

[mmm4444bot]

- 13/34 (-0.3823)

As an estimate, that's pretty good.

You can find a better answer (rounded to four places) by not rounding your intermediate results so much. (We call this round-off error).

Keep six decimal places, as you work. Then you can be sure that your rounded answer is precise.

:idea: If your calculator has a storage feature for results, then let the calculator carry all of the digits for you, and you round only the final result.

Does that mean you just change it to positive. (e.g. = 0.3823 metres after two seconds)

Not quite. The units on velocity are metres per second (m/s). The negative sign tells us that the object moved "backwards".

Also, the average value that you found is not the velocity after two seconds; it's an average velocity over the first two seconds of movement.

So, we can say the answer is:

Avg Velocity = -0.3946 m/s

This means:

The avg speed of the particle was 0.3946 m/s AND it was moving in the negative direction.

If your course has mentioned vectors, then look up the definitions for "vector" and "scalar", and compare them.

Velocity is a vector quantity; it has both magnitude and direction.

Speed is a scalar quantity; it has no direction; it's the absolute value of the vector quantity.

When we make statements and direction does not matter, we would use the speed; when we do care about the direction, as well, then we would use the velocity.

They asked for the average velocity over the first two seconds, so we report the vector -0.3946 m/s. :cool:

 

[tkhunny]

Don't confuse "Speed" and "Velocity".