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# Answer to Question #128866 in Mechanics | Relativity for Ariyo Emmanuel

Question #128866
At the instant shown in Fig. 1, cars A and B are traveling with speeds of 18 m/s and 12 m/s, respectively. Also at this instant, A has a decrease in speed of 2 m/s2, and B has an increase in speed of 3 m/s2
velocity and acceleration of B with respect to A.
1
2020-08-24T13:04:50-0400

Solution

Given data

speed of car A = 18 m/s

speed of car B = 12m/s

we are assuming both cars are moving along +x-axis direction then velocity of car A (VA)= +18m/s

velocity of car B (VB)=+12 m/s

acceleration of car A (aA)= -2 m/s^2

acceleration of car B (aB)= +3 m/s^2

Now by apply of fomula of relative velocity

VBA=VB -VA

by putting the values of VA and VB

VBA=12-18

VBA= -6 m/s

it mean velocity of B with respect to A is along -X axis direction.

By the formula of relative acceleration of car B with

car A is

aBA=aB - aA

=3 - (-2)

aBA =5 m/s^2

it mean acceleration of B with respect to A is along +X axis direction.

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