Car A rounds a curve of 150‐m radius at a constant speed of 54 km/h. At the instant represented, car B is moving at 81 km/h but is slowing down at the rate of 3 m/s2. Determine the velocity and acceleration of car A as observed from car B.

Velocity of Afrom B=21m/s

Acceleration of A from B=1.68m/s°2

Explanation:

Given

Radius r=150m

Velocity of a Va= 54km/hr

Va=54*1000/3600=15m/s

Velocity of b Vb=82km/hr

VB=81*1000/3600=22.5mls

The velocity of Car A as observed from B is VBA

VB= VA+VBA

Resolving the vector into X and Y components

For X component= 15cos60=7.5m/s

Y component=22 5sin60=19.48m/s

VBA= √(X^2+Y^2)

VBA= ✓(7.5^2+19.48^2)=21m/s

For acceleration of A observed from B

A=VA^2/r= 15^2/150=1.5m/s

Resolving into Xcomponent=1.5cos60=0.75m/s

Y component=3cos60=1.5

Acceleration BA=√(0.75^2+1.5^2)

1.68m/s

trig 15.3 cos

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.