Question

An object has a position given by r = [2.0 m + (2.00 m/s)t] i + [3.0 m − (1.00 m/s^2)t^2] j, where quantities are in SI units. What is the speed of the object at time t = 2.00 s?

Answer 1

Answer: 1 m/s

Explanation:

We have an object whose position $r$ is given by a vector, where the components X and Y are identified by the unit vectors $i$ and $j$ (where each unit vector is defined to have a magnitude of exactly one):

$r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j$

On the other hand, velocity is defined as the variation of the position in time:

$V=\frac{dr}{dt}$

This means we have to derive $r$:

$\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j$

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} t) j$ This is the velocity vector

And when $t=2s$ the velocity vector is:

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} (2 s)) j$

$\frac{dr}{dt}=2 m/s i - 1m/s j$ This is the velocity vector at 2 seconds

However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed $S$:

$S=\sqrt {-1 m/s j + 2 m/s i}$

$S=\sqrt {1 m/s}$

Finally:

$S=1 m/s$ This is the speed of the object at 2 seconds