Question

A car a is going north east at 80km/hr and a car b is going at 60 km/hr then the direction of velocity of a with respect to b makes an anglle alpha with north such that tan alpha is

Answer 1

velocity of a with respect to b makes an anglle alpha with north such that tan alpha is 1/7.

vector form of velocity of car A, $v_A$ = 80cos45° i + 80sin45° j

= 40√2 i + 40√2 j

vector form of velocity of car B, $v_B$ = 60cos45° i - 60sin45° j

= 30√2 i - 30√2 j

now velocity of car A with respect to car B , $v_{AB}=v_A-v_B$

=(40√2 i + 40√2 j) - (30√2 i - 30√2 j)

= 10√2 i + 70√2 j

so, angle made by $v_{AB}$ with north, α = tan-¹(x - component of $v_{AB}$/y - component of $v_{AB}$ )

= tan-¹(10√2/70√2)

= tan-¹(1/7)

so, tanα = 1/7

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Answer 2

$\huge\bold\purple{Answer:-}$

vector form of velocity of car A, v_A = 80cos45° i + 80sin45° j

= 40√2 i + 40√2 j

vector form of velocity of car B, v_B = 60cos45° i - 60sin45° j

= 30√2 i - 30√2 j

now velocity of car A with respect to car B , v_{AB}=v_A-v_B

=(40√2 i + 40√2 j) - (30√2 i - 30√2 j)

= 10√2 i + 70√2 j

so, angle made by v_{AB} with north, α = tan-¹(x - component of v_{AB}/y - component of v_{AB} )

= tan-¹(10√2/70√2)

= tan-¹(1/7)

so, tanα = 1/7