A particle's position along the x-axis is described by: x(t)=A t + B t2
1 answer:
Answer:
a)V = 25.1 m/s
b)V = 4.226 m/s
Explanation:
Given that
x(t)=A t + B t²
A = -4.3 m/s
B = 4.9 m/s²
x(t)= - 4.3 t +4.9 t²
The velocity of the particle is given as
V=-4.3 + 4.9 x 2 t
V= - 4.3 + 9.8 t m/s
Velocity at point t= 3 s
V= - 4. 3 + 9.8 x 3 m/s
V= - 4.3 + 29 .4 m/s
V = 25.1 m/s
At origin :
x= 0 m
0 = - 4.3 t +4.9 t²
0 = - 4.3 + 4.9 t
t=0.87 s
The velocity at t= 0.87 s
V= - 4.3 + 9.8 t m/s
V= - 4. 3 + 9.8 x 0.87 m/s
V= - 4.3 + 8.526 m/s
V = 4.226 m/s
a)V = 25.1 m/s
b)V = 4.226 m/s
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Given,
The momentum of the object A before the collision, p₁ =80 Ns
The momentum of the object B before the collision, p₂=-30 Ns
Given that the objects stick together after the collision.
From the law of conservation of momentum, the total momentum of a system should always remain the same. Thus the total momentum of the objects before the collision must be equal to the total momentum of the objects after the collision.
Thus,
Where p is the total momentum of the system at any instant of time.
On substituting the known values,
Therefore the total momentum of the system is 50 Ns
Thus the momentum of the object AB after the collision is 50 Ns