Question

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.

Answer 1

Velocity of Afrom B=21m/s

Acceleration of A from B=1.68m/s°2

Step-by-step 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

Answer 2

The velocity of the car A as observed from car B is (15i - 22.5j) m/s

The acceleration of car A as observed from car B is 4.5j m/s²

Step-by-step explanation:

The velocity of car A and car B in m/s is equal:

`v_(A) =54(km)/(h) *(5)/(18) =15im/s`

`v_(B) =81(km)/(h) *(5)/(18) =22.5jm/s`

The relative velocity is:

`v_(AB) =v_(A) -v_(B) =(15i-22.5j)m/s`

The acceleration of the cars are:

`a_(A) =(v_(A)^(2) )/(r) =(15^(2) )/(150) =1.5jm/s^(2)`

The relative acceleration is:

`a_(AB) =a_(A) -a_(B) =1.5-(-3),negative-because-the-car-is-slowing-down\\a_(AB)=4.5jm/s^(2)`