1. Jan 25, 2008

### Go Angels

Suppose that the position-vector function 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 = 1.40 m/s, b = 0.75 m, c = 0.120 m/s2, and d = 1.10 m.

(a) Calculate the average velocity during the time interval from t = 1.75 s to t = 3.85 s.
( ___ î + ___ ĵ ) m/s

(b) Determine the velocity at t = 1.75 s.
( ___ î + ___ ĵ ) m/s

Determine the speed at t = 1.75 s.
___ m/s

The answers I got for the first four slots are: 3.2, 1.46, 1.4, and 0.42. I don't know if those values are correct. And can someone please teach me how to do part 5? Thank you.

x(t) = 1.4(t) + 0.75
y(t) = 0.12(t)^2 +1.1

2. Jan 25, 2008

### physixguru

Substitute the value of t as 1.75 & differentiate the equation wid respect to time and then ya get the velocity....

3. Jan 25, 2008

### Go Angels

So for part a...
1.4(1.75) + 0.75 = 3.2
&
0.12(1.75)^2 +1.1 = 1.46

For part b, after differentiating I get...
1.4
&
0.24(1.75) = 0.42

Is that correct?

The part I'm most confused about is the finding the speed for the last part. I know it has to do with magnitude.

4. Jan 25, 2008

### physixguru

ur correct...now take modulus of velocity to get the speed....or in other words magnitude as ya say...

5. Jan 25, 2008

### Go Angels

Opps, for part a.) would I have to divide those values by 2.1? Since it says during the time interval betw. t = 1.75 s to t = 3.85 s.

And can you describe to me how to go about finding the magnitude please?

6. Jan 25, 2008

### physixguru

of course u would have to divide it by 2.1...

now for magnitude..

magnitude i.e.speed= {[velocity along x axis]^2+ [velocity along y-axis]^2}^1/2

7. Jan 25, 2008

### Go Angels

Thank you very much