Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
What does this mean? What do you need help with?
Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:
(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>
where <em>v</em> is the velocity of the system. Solve for <em>v</em> :
<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)
<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s
Knowing the direction of a force is important because it helps someone know the motion of the object. if you use a free body diagram, then it becomes easy to see all the forces being applied to an object. if there is more force going one way, the object is accelerating in that direction. if all the forces cancel each other out, then the object is at a constant speed or is at rest.
