Vectors and Motion: Calculating Average Velocity and Speed

[Husker70]

Homework Statement

Suppose the positon vector for a particle is given as a function of time by vector r(t)
=x(t)i + y(t)j with x(t) =at + b and y(t) = ct2 + d, where a=1.00 m/s, b=1.00m,
c=.125 m/s2, and d=1.00m. (a) Calculate the average velocity from t=2.00s to t=4.00s.
(b) Determine the velocity and the speed at t=2.00s

Homework Equations

(a) plug in the values for a, b, c, and d to
x(t) = at + b and y(t) = ct2 + d
Vavg = delta r / delta t


3. The Attempt at a Solution [/c]
(a) Ri =
x(t) = (1.00m/s) 2.00s + 1.00m = 3.00 m/s
y(t) = (.125m/s2) 2.00s2 + 1.00m = 2.41 m/s
Rf
x(t) = (1.00m/s) 4.00s + 1.00m = 5m/s
y(t) = (.125m/s) 4.00s2 + 1.00m = 6.66m/s

Rf -Ri / Tf - Ti

Rf = 6.66 - 2.41 = 4.25
Ri = 5 - 3 = 2
t = 2
4.25 - 2 /2 = 1.13 Vavg from t = 2 to t=4

(b) From equation at 2 s
velocity is
x(t) = 2.41
y(t) = 3
velocity is 5.41 m/s at t=2

Speed is magnitute of vector square root of (3)2 + (2.41)2 = square root of 14.81
Speed = 3.85 m/s

Am I right or close?
Thanks for the help

Kevin

 

[LowlyPion]

Homework Statement

Suppose the positon vector for a particle is given as a function of time by vector r(t)
=x(t)i + y(t)j with x(t) =at + b and y(t) = ct2 + d, where a=1.00 m/s, b=1.00m,
c=.125 m/s2, and d=1.00m. (a) Calculate the average velocity from t=2.00s to t=4.00s.
(b) Determine the velocity and the speed at t=2.00s

Homework Equations

(a) plug in the values for a, b, c, and d to
x(t) = at + b and y(t) = ct2 + d
Vavg = delta r / delta t


3. The Attempt at a Solution [/c]
(a) Ri =
x(t) = (1.00m/s) 2.00s + 1.00m = 3.00 m/s
y(t) = (.125m/s2) 2.00s2 + 1.00m = 2.41 m/s
Rf
x(t) = (1.00m/s) 4.00s + 1.00m = 5m/s
y(t) = (.125m/s) 4.00s2 + 1.00m = 6.66m/s

Rf -Ri / Tf - Ti

Rf = 6.66 - 2.41 = 4.25
Ri = 5 - 3 = 2
t = 2
4.25 - 2 /2 = 1.13 Vavg from t = 2 to t=4

(b) From equation at 2 s
velocity is
x(t) = 2.41
y(t) = 3
velocity is 5.41 m/s at t=2

Speed is magnitute of vector square root of (3)2 + (2.41)2 = square root of 14.81
Speed = 3.85 m/s

Am I right or close?
Thanks for the help

Kevin

Without checking the actual math, it looks like you have grasped the concepts.

Cheers.

 

[alphysicist]

Homework Statement

Suppose the positon vector for a particle is given as a function of time by vector r(t)
=x(t)i + y(t)j with x(t) =at + b and y(t) = ct2 + d, where a=1.00 m/s, b=1.00m,
c=.125 m/s2, and d=1.00m. (a) Calculate the average velocity from t=2.00s to t=4.00s.
(b) Determine the velocity and the speed at t=2.00s

Homework Equations

(a) plug in the values for a, b, c, and d to
x(t) = at + b and y(t) = ct2 + d
Vavg = delta r / delta t


3. The Attempt at a Solution [/c]
(a) Ri =
x(t) = (1.00m/s) 2.00s + 1.00m = 3.00 m/s
y(t) = (.125m/s2) 2.00s2 + 1.00m = 2.41 m/s

This equation means

$$y(t) = (.125m/s^2) (2.00s)^2 + 1.00m$$
right? I don't believe the answer is 2.41. It looks like the same things happens for y(t) for the Rf value.

Rf
x(t) = (1.00m/s) 4.00s + 1.00m = 5m/s
y(t) = (.125m/s) 4.00s2 + 1.00m = 6.66m/s

Rf -Ri / Tf - Ti

Rf = 6.66 - 2.41 = 4.25
Ri = 5 - 3 = 2

These last two equations do not match your formula. The value 2 is the change in the x component of the displacement, and the 4.25 (though that number does not look right to me from above) is the change in the y component of the displacement.

That is (using the value of 4.25 for the moment),

$$\Delta \vec R = \vec R_f - \vec R_i = 2 \hat i + 4.25 \hat j$$

which is what you would use in your average velocity equation:

$$\vec v_{\rm ave} = \frac{\Delta \vec R}{\Delta t}$$

t = 2
4.25 - 2 /2 = 1.13 Vavg from t = 2 to t=4

(b) From equation at 2 s
velocity is
x(t) = 2.41
y(t) = 3
velocity is 5.41 m/s at t=2

The values of 3 and 2.41 were the positions you found at t=2s (but they were the other way around; you had x=3 and y=2.41). But you need to find the components of the velocity at t=2 seconds.

Based on the x(t) and y(t) equation they give you, what are the equations for [tex]v_x(t)[/itex] and $v_y(t)$?<br /> <br /> (Once you have the velocity components, you cannot just add the numerical values together to get the velocity, since they are not in the same direction.)<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Speed is magnitute of vector square root of (3)2 + (2.41)2 = square root of 14.81<br /> Speed = 3.85 m/s </div> </div> </blockquote><br /> Once you find the right components of the velocity, this is the way to find the speed.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Am I right or close?<br /> Thanks for the help<br /> <br /> Kevin </div> </div> </blockquote>[/tex]

 

[LowlyPion]

My apologies. I looked briefly at the method and thought you were substituting values and calculating correctly using the RSS to get your resultant magnitude. I should have actually checked your math more closely now I see.

Thanks to alphysicist for catching the errors here.