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

Question

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.

Answer 1

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 (V_A)= +18m/s

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

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

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

Now by apply of fomula of relative velocity

V_BA=V_B -V_A

by putting the values of V_A and V_B

V_BA=12-18

V_BA= -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

a_BA=a_B - a_A

_{=3 - (-2)}

a_{BA =5 m/s^2}

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