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sp2606 [1]
3 years ago
13

If Maggie is walking at 1.4 m/s and accelerates at 0.20 m/s^2. What is her final velocity at the end of 100.0m

Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

6.478m/s

Explanation:

There are some assumptions to be made. To answer this problem, I assume that Maggie is walking in a straight line and that she initially starts off with a velocity of 1.4m/s.

Maggie initially walks at a speed of 1.4m/s. We are also told that the acceleration is 0.2m/s^2 and that she walked a total of 100m. Therefore our parameters are:

v_i=1.4\frac{m}{s}\\a=0.2\frac{m}{s^2}\\d=100m\\v_f=?

Looking up kinematic equations that contains these parameters are

v_f^2=v_i^2+2ad\\v_f^2=(1.4\frac{m}{s})^2+2*(0.2\frac{m}{s^2})(100m)\\v_f^2=41.96\frac{m^2}{s^2}\\v_f=6.478\frac{m}{s}

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You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 15.4 m above the ground, i
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I believe there are two things which we asked to find here:

1. its speed just as it left the ground

2. find its maximum height

 

Solutions:

1. We use the formula:

ΔKE = - ΔPE

where KE is kinetic energy = ½ mv^2, and PE is potential energy = m g h, Δ = change

Therefore:

½ m (v2^2 – v1^2) = m g (h1 – h2)

at initial point, point 1: h1 = 0, v1 = ?

at final point, point 2: h2 = 15.4 m, v2 = 24.2 m/s

½ (24.2^2 – v1^2) = 9.8 (0 – 15.4)

585.64 – v1^2 = -301.84

-v1^2 = 887.48

v1 = 29.8 m/s

So the rock was travelling at 29.8 m/s as it left the ground.

 

2. The maximum height (hmax) reached is calculated using the formula:

v1^2 = 2 g hmax

Rewriting in terms of hmax:

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Therefore the rock reached a maximum height of 45.3 meters.

4 0
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