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Car A has an acceleration of $2\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}$ due east and car B $4\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}$ due north. What is the acceleration of car B with respect to car A?

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: The given case is in a two dimensional system, therefore the directional vectors should also be considered. The total magnitude of the acceleration depends on the acceleration among different directions.

Complete step by step answer:
When the two cars are moving in the different directions then the relative velocity of the cars is calculated by subtracting the velocities of the car. When the car moves in directly opposite directions then the relative velocity of the car is calculated by the sum of two velocities.
The direction of the cars is as shown:
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The formula to calculate the relative acceleration of car B with respect to car A is
${a_{BA}} = {a_B}\hat j - {a_A}\hat i$
Here, ${a_{BA}}$ is the relative acceleration of car A with respect to car B, ${a_A}$ is the acceleration of car A, $\hat i$ is the unit vector along east direction, ${a_B}$is the acceleration of car B and $\hat j$ is the unit vector along north direction.
Substitute $2\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.
 } {{{\rm{s}}^{\rm{2}}}}}$ for ${a_A}$ and $4\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}$for ${a_B}$ in the formula to calculate the relative acceleration of the car B with respect to car A.
${a_{BA}} = \left( {2\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\hat j - \left( {4\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\hat i$
The formula to calculate the magnitude of relative acceleration of car B with respect to car A is
$\left| {{a_{BA}}} \right| = \sqrt {a_B^2 + a_A^2} \,\,$
Substitute $2\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}$ for ${a_A}$ and $4\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}$for ${a_B}$ in the formula to calculate magnitude of the relative acceleration of the car B with respect to car A.
$ \left| {{a_{BA}}} \right| = \sqrt {{{\left( {2\,{{\rm{m}} {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}} \right)}^2} + {{\left( {4\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}^2}} \\
 = \sqrt {4 + 16} \,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}\\
 = \sqrt {20} \,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\\
 = 4.47\,{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}} $
Thus, the relative acceleration of car B with respect to car A is $4.47\,{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}$.

Note:
The direction of car B relative to car A can also be found from this example because the direction of the car can be calculated using the acceleration of the cars in different directions.