Question

Lisa in her Lamborghini accelerates at the rate of $(3.00\overrightarrow i - 2.00\overrightarrow j )m/{s^2}$ , while Jill in her Jaguar accelerates at $(1.00{\overrightarrow i ^{}} + 3.00\overrightarrow j )m/{s^2}$ . They both start from rest at the origin of an xy coordinate system. After $5s$ , (a) what is Lisa’s speed with respect to Jill, (b) how far apart are they, and (c) what is Lisa’s acceleration relative to Jill?

Answer 1

To solve questions involving more than one dimension, we must have clear ideas of coordinates and directions. It is even better that we draw the coordinate system and solve the problem. Moreover, this problem involves relative motion, we have to understand the relative motion concept also.

Complete step by step answer:
Given the acceleration of Lisa and Jill respectively are :
Acceleration of Lisa( ${\overrightarrow a _L}$ ) = $(3.00\overrightarrow i - 2.00\overrightarrow j )m/{s^2}$
Acceleration of Jill ( ${\overrightarrow a _J}$ ) = $(1.00{\overrightarrow i ^{}} + 3.00\overrightarrow j )m/{s^2}$

Now in-order to get the speed we have to integrate the value of acceleration, as we know that acceleration is the rate of change of velocity. Now it is also given that the value of velocity at the start is zero, as it is at rest hence there’s no constant of integration.
Thus velocity of Lisa ( ${\overrightarrow v _L}$ ) = $(3.00\widehat {ti} - 2.00\widehat {tj})m/{s^2}$.
Velocity of Jill ( ${\overrightarrow v _J}$ ) = $(1.00\widehat {ti} + 3.00\widehat {tj})m/{s^2}$

Now let’s find the value of position. To find that we integrate the velocity, and now too the constant of integration is zero as it is already given in question that they are at origin at the beginning.
Thus position of Lisa ( ${\overrightarrow r _L}$ ) = $1.50{t^2}\widehat i - 1.00{t^2}\widehat j$
Position of Jill ( ${\overrightarrow r _J}$ ) = $0.50{t^2}\widehat i + 1.50{t^2}\widehat j$

Now these are given with respect to the ground, and it asks in the question, Lisa’s velocity with respect to Jill, thus
${\overrightarrow v _{LJ}} = {\overrightarrow v _L} - {\overrightarrow v _J} = (3.00\widehat {ti} - 2.00\widehat {tj})m/{s^{}} - (1.00\widehat {ti} + 3.00\widehat {tj})m/{s^{}}$
Now after the interval of $5$ second, velocity will be = $(10.00\widehat i - 25.00\widehat j)m/{s^{}}$
The magnitude: $\sqrt {{{10}^2} + {{25}^2}} = 26.9m/{s^2}$

(b) Now we have to find the position of Lisa with respect to Jill in-order to get the distance between them after $5$ seconds.Thus
${\overrightarrow r _{LJ}}$= ${\overrightarrow r _L}$ - ${\overrightarrow r _J}$
= $1.50{t^2}\widehat i - 1.00{t^2}\widehat j$ - ( $0.50{t^2}\widehat i + 1.50{t^2}\widehat j$ )
Thus after $5$ seconds position is : $25\widehat i - {62.5^{}}\widehat j$
The magnitude = $67.3$ m .

(c) The relative velocity between Lisa and Jill is :
${\overrightarrow a _{LJ}}$ = $(3.00\overrightarrow i - 2.00\overrightarrow j )m/{s^2}$ - $(1.00{\overrightarrow i ^{}} + 3.00\overrightarrow j )m/{s^2}$
Thus, the acceleration is: $(2.00\widehat i - 5.00\widehat j)m/{s^2}$
The magnitude: $\sqrt {{2^2} + {5^2}}$ = $5.385$ $m/{s^2}$

Note: These questions need to be understood by the concept of directions and the value is obtained in both the coordinates. To get the magnitude, we simply need to square the values of each coordinate and add them, then take the square root of the sum.Remember that always values given such as velocity and acceleration are in respect to the ground, unless mentioned in the question.