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77julia77 [94]
3 years ago
8

A truck initially traveling at a speed of 22 meters per second increases speed at a constant rate of 2.4 meters per second^2 for

3.2 seconds. what is the total distance traveled by the truck during this 3.2 second time interval
Physics
1 answer:
Usimov [2.4K]3 years ago
8 0
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:

d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m 
Hope the answer helps. 
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Two carts, A and B, are connected by a spring and sitting at rest on a track. Cart A has a mass of 0.4 kg and Cart B has a mass
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Both carts experience the same force but Cart A has a greater speed after the recoil.

The given parameters;

  • <em>Mass of the cart A = 0.4 kg</em>
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Apply the principle of conservation of linear momentum to determine the velocity of the carts after collision;

m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.

F_A = -F_B

Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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