Respuesta :
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:
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:
[tex]v_{A} =54\frac{km}{h} *\frac{5}{18} =15im/s[/tex]
[tex]v_{B} =81\frac{km}{h} *\frac{5}{18} =22.5jm/s[/tex]
The relative velocity is:
[tex]v_{AB} =v_{A} -v_{B} =(15i-22.5j)m/s[/tex]
The acceleration of the cars are:
[tex]a_{A} =\frac{v_{A}^{2} }{r} =\frac{15^{2} }{150} =1.5jm/s^{2}[/tex]
The relative acceleration is:
[tex]a_{AB} =a_{A} -a_{B} =1.5-(-3),negative-because-the-car-is-slowing-down\\a_{AB}=4.5jm/s^{2}[/tex]
