Objects A and B are at the origin. If object A moves to #(0 ,4 )# and object B moves to #(9 ,8 )# over #3 s#, what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

Question

Objects A and B are at the origin.  If object A moves to #(0  ,4  )# and object B moves to #(9  ,8  )# over #3  s#, what is the relative velocity of object B from the perspective of object A?  Assume that all units are denominated in meters.

Elena Albarran · Accepted Answer

#3.14# #"m/s"#

We're asked to find the relative velocity of object #B# with respect to object #A#, given their positions after a time interval.

Let's first find the (assumed to be constant) velocity components of #A# and #B# with respect to the origin:

#v_(Ax"/"O) = (4"m")/(3"s") = 1.33# #"m/s"#

#v_(Ay"/"O) = (0"m")/(3"s") = 0# #"m/s"#

#v_(Bx"/"O) = (9"m")/(3"s") = 3# #"m/s"#

#v_(By"/"O) = (8"m")/(3"s") = 2.67# #"m/s"#

The equation here for the velocity of #B# with respect to #A# is given by

#v_(B"/"A) = v_(B"/"O) + v_(O"/"A#

Here, the velocity #v_(O"/"A# of the origin with respect to #A# is simply the negative of the velocity of #v_(A"/"O)#

Dissecting this into its constituent parts, we have

#v_(Bx"/"Ax) = v_(Bx"/"O) + v_(O"/"Ax)#

#v_(By"/"Ay) = v_(By"/"O) + v_(O"/"Ay)#

Thus,

#v_(Bx"/"Ax) = 3"m/s" + (-1.33"m/s") = 1.67# #"m/s"#

#v_(By"/"Ay) = 2.67"m/s" + (-0"m/s") = 2.67# #"m/s"#

The magnitude of #vecv_(B"/"A)# is given by

#v_(B"/"A) = sqrt((v_(Bx"/"Ax))^2 + (v_(By"/"Ay))^2)#

#= sqrt((1.67"m/s")^2 + (2.67"m/s")^2) = color(red)(3.14# #color(red)("m/s"#

Thus, the relative speed of object #B# relative to #A# is #3.14# meters per second.

Lily Adams · Answer

undefined To find the relative velocity of object B from the perspective of object A, we need to calculate the displacement of B relative to A over the given time interval.

The displacement of B relative to A is given by the final position of B minus the initial position of B, subtracted by the final position of A minus the initial position of A.

Let's denote the position of object A as $ (x_{A}, y_{A}) $ and the position of object B as $ (x_{B}, y_{B}) $.

The initial position of A is $ (0, 0) $, the final position of A is $ (0, 4) $.
The initial position of B is $ (0, 0) $, the final position of B is $ (9, 8) $.

Using these values, we can calculate the displacement of B relative to A:

$$
\Delta x_{BA} = x_{B} - x_{A} = 9 - 0 = 9 \text{ meters}
$$
$$
\Delta y_{BA} = y_{B} - y_{A} = 8 - 4 = 4 \text{ meters}
$$

Now, we can calculate the relative velocity of B from the perspective of A by dividing the displacement by the time interval:

$$
v_{BAx} = \frac{\Delta x_{BA}}{3 \text{ s}} = \frac{9}{3} = 3 \text{ m/s}
$$
$$
v_{BAy} = \frac{\Delta y_{BA}}{3 \text{ s}} = \frac{4}{3} = \frac{4}{3} \text{ m/s}
$$

Therefore, the relative velocity of object B from the perspective of object A is $ (3 \text{ m/s}, \frac{4}{3} \text{ m/s}) $.