Help with velocity and acceleration

[skp123]

The 400 kg mine car is hoisted up the incline with a slope
angle of 30°. The massless cable connects the mine car and
the motor that induces its motion. The force in the cable is
defined by following function F = 3200 t2 [N], where the
time t is in seconds. The mine car has at t = 0 and s = 0 an
initial velocity v1 = 2 m/s. There is no friction between the
mine car and the incline.

a) Determine the acceleration of the mine car for t = 1 s.
b) Calculate the velocity of the mine car when t = 2 s.
c) Determine the position of the mine car on the plane after t = 3 s.

Can anyone help me to solve this problem? Just tell me which formulas to use in order to solve this problem .

 

[jbsteele]

Thanks in advance. A) The acceleration of the mine car for t = 1 s can be determined using Newton's Second Law: F = ma. Thus, a = F/m = 3200 t2/400 kg = 8 m/s2. B) The velocity of the mine car when t = 2 s can be calculated using the equation v2 = v1 + at, where v1 is the initial velocity and a is the acceleration. Thus, v2 = 2 m/s + 8 m/s2 x 2 s = 20 m/s. C) The position of the mine car on the plane after t = 3 s can be determined using the equation s = v1t + 1/2at2, where v1 is the initial velocity, a is the acceleration, and t is the time. Thus, s = 2 m/s x 3 s + 1/2 x 8 m/s2 x (3 s)2 = 44 m.

 

[kerimek]

Sure, I would be happy to help you with this problem. In order to solve this problem, we will need to use the equations of motion for objects undergoing constant acceleration. These equations are:

1) v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2) s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

3) v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

a) To determine the acceleration of the mine car at t = 1 s, we can use equation 1. We know that the initial velocity is 2 m/s and the time is 1 s, so we have:

v = u + at
a = (v-u)/t
a = (0-2)/1 = -2 m/s^2

Therefore, the acceleration of the mine car at t = 1 s is -2 m/s^2.

b) To calculate the velocity of the mine car at t = 2 s, we can use equation 1 again. This time, we know that the initial velocity is 2 m/s, the time is 2 s, and we have just calculated the acceleration to be -2 m/s^2. So we have:

v = u + at
v = 2 + (-2)(2)
v = 2 - 4 = -2 m/s

Therefore, the velocity of the mine car at t = 2 s is -2 m/s.

c) To determine the position of the mine car on the plane after t = 3 s, we can use equation 2. We know that the initial velocity is 2 m/s, the time is 3 s, and the acceleration is -2 m/s^2. So we have:

s = ut + 1/2at^2
s = (2)(3) + 1/2(-2)(3)^2
s = 6 - 9 = -3 m

Therefore, the position of the mine car on the plane after t = 3 s is -3 m.

I hope this helps you to solve the problem. Let me