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

Answer:

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

Answer 2

Answer:

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

Explanation:

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

$v_{A} =54\frac{km}{h} *\frac{5}{18} =15im/s$

$v_{B} =81\frac{km}{h} *\frac{5}{18} =22.5jm/s$

The relative velocity is:

$v_{AB} =v_{A} -v_{B} =(15i-22.5j)m/s$

The acceleration of the cars are:

$a_{A} =\frac{v_{A}^{2} }{r} =\frac{15^{2} }{150} =1.5jm/s^{2}$

The relative acceleration is:

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