Ehh, Earth rotates around its axis and Earth revolves around the sun as it travels through space.
<span>(a) -9.97 m/s
(b) x = 2.83
This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.
f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2
So the acceleration of the body is now expressed as
f''(x) = -2.580645161x m/s^2
Let's calculate the anti-derivative from that.
f''(x) = -2.580645161x m/s^2
f'(x) = -1.290322581x^2 + C m/s
Now let's use the known velocity value at x = 2.0 to calculate C
f'(x) = -1.290322581x^2 + C
1
1 = -1.290322581*2^2 + C
11 = -1.290322581*4 + C
11 = -5.161290323 + C
16.161290323 = C
So the velocity function is
f'(x) = -1.290322581x^2 + 16.161290323
(a) The velocity at x = 4.5
f'(x) = -1.290322581x^2 + 16.161290323
f'(4.5) = -1.290322581*4.5^2 + 16.161290323
f'(4.5) = -1.290322581*20.25 + 16.161290323
f'(4.5) = -26.12903227 + 16.161290323
f'(4.5) = -9.967741942
So the velocity is -9.97 m/s
(b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
0 = -1.290322581x^2 + 10.36129032
1.290322581x^2 = 10.36129032
x^2 = 8.029999998
x = 2.833725463</span>
60 miles / hr
= 60 * 1.6 * 1000 m / 60 * 60 s
= 96000 m / 3600 s
= 26.67 m / s
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