5. Graph A plots a race car's speed for 5 seconds. The car's
1 answer:
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Answer:
a) v² = v₀² + 2 g h, b) t = v₀/g (1+ √ (1 + 2gh/ v₀²))
Explanation:
a) This is an exercise that we can solve using conservation of energy.
Starting point. High point
Em₀ = K + U = ½ m v₀² + m gh
Final point. Soil
= K = ½ m v²
energy is conserved because there is no friction
Emo = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
b) the time it takes to reach the ground can be calculated with kinematics
let's create a reference frame with positive upward direction
v = vo - g t
when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀
v = -v₀ - gt
t = - (v + v₀) / g
we substitute the velocity values calculated in the previous part
t = - (√(v₀² + 2 g h) + vo) / g
we will simplify the equation a bit
t = - v₀/g (1+ √ (1 + 2gh/ v₀²))
c) is now thrown vertically upward with the same initial velocity vo.
To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a
v = √ (v₀² + 2gh)
d) for this part if there is change since the speed is not squared
v₀ = v₀
v = v₀ - gt
t = (v₀ - v) / g
t = (v₀ - √(v₀² + 2 g h)) / g
t = v₀/g (1 - √(1 + 2gh / v₀²))