# Given the following formulas

## Homework Statement

Suppose that the position vector for a particle is given as a function of time by (t) = x(t) + y(t), with x(t) = at + b and y(t) = ct2 + d,where a = 1.70 m/s, b = 1.20 m, c = 0.122 m/s2, and d = 1.18 m.
(a) Calculate the average velocity during the time interval from t = 2.05 s to t = 4.05 s.
(b)Determine the velocity at t = 2.05 s.
(c)Determine the speed at t = 2.05 s.

## Homework Equations

for a.) I used Vavg=delta r/delta t

## The Attempt at a Solution

for a.) for rf i got x(t)=8.05
y(t)=3.181

for ri i got x(t)=4.685
y(t)=1.69

and delta t=4.05-2.05=2

Last edited:

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rl.bhat
Homework Helper
Post the complete problem.

listing the variables as a,b,c,d does us no good. We can assume what they are with the units but i don't know what b and d are.

Basically, use the standard symbols.

Simon Bridge
Homework Helper
Do I take it that: $$\vec{r}(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}$$... where: $x(t)=v_xt+x_0$ and $y(t)=a_yt^2+y_0$

These are position-time functions - I changed the variable names to reflect their roles.
From these you can find $v_x(t)$, and $v_y(t)$

(a) average velocity is change in position over change in time - you have two position vectors - what is the distance between their end-points?

(b) $v_x(t)=\dot{x}(t)$

(c) how is the speed and the velocity related?

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Suppose that the position vector for a particle is given as a function of time by (t) = x(t) + y(t), with x(t) = at + b and y(t) = ct2 + d,where a = 1.70 m/s, b = 1.20 m, c = 0.122 m/s2, and d = 1.18 m.
(a) Calculate the average velocity during the time interval from t = 2.05 s to t = 4.05 s.
(b)Determine the velocity at t = 2.05 s.
(c)Determine the speed at t = 2.05 s.

## Homework Equations

for a.) I used Vavg=delta r/delta t

## The Attempt at a Solution

for a.) for rf i got x(t)=8.05
y(t)=3.181

for ri i got x(t)=4.685
y(t)=1.69

and delta t=4.05-2.05=2

Do you know Calculus?

Simon Bridge
Homework Helper
Gudduy cobber;
Does not need calculus if kinematics is available
... anyway, piercegirl got stuck on average velocity. The calc is in the next bit.

I know - calculus is fun and we want to spread the Word... ;)

I like Serena
Homework Helper
Hi piercegirl!

You're on the right track.
I guess you need to be aware of the fact that an average "velocity" is a vector.

You have position vectors at both locations.
Can you subtract those position vectors, yielding a change in position vector?
And divide that by the change in time?

In other words, can you calculate:
$$\mathbf{\vec v}_{avg} = {\mathbf{\vec{Δr}} \over Δt}$$