Finding velocity vector from position change at constant speed

[Gughanath]

On an occasion, John has position vector 40j metres. He wished to skate in a straightline to the point with position vector 30i metres. Given that his speed is constant at 5ms^-1, find his velocity.
Someone help me on this one please!

 

[BobG]

You have a magnitude (5 meters per second).

The starting coordinates (0,40) and the ending coordinates (30,0) supply the direction. Relative to your origin (0,0) your two position vectors are 0i + 40j and 30i + 0j

Since you're using ijk coordinates, your best bet is to add your two position vectors together. Then you have a scalar problem: x(30i + 40j) = v

To find x, you have to divide the magnitude of the velocity vector by the magnitude of the resultant position vector. Plug the coordinates of your position vector into the Pythagorean Theorem.

Once you determine x, do the scalar mutliplication (multiply each of the resultant position vector coordinates by x).

 

[Gughanath]

Oh!...Thanx

 

[OlderDan]

Oh!...Thanx

You need the difference between the final and initial position vectors, not their sum, to get the direction of your velocity vector

$$\overrightarrow {\Delta r} = \overrightarrow{r}_{final} - \overrightarrow {r}_{initial} = 30\widehat{i} - 40\widehat{j}$$

Since you know the magnitude of your velocity vector, you need to find the unit vector in the direction of $$\overrightarrow {\Delta r}$$. You can do that by dividing the vector $$\overrightarrow {\Delta r}$$ by its length. Then multiply that unit vector by the magnitude of the velocity you have.

 

[BobG]

Oops. OlderDan's right. It's final position minus start position.

It's -30i +40j , not +30i.