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Karolina [17]
3 years ago
9

Suppose you increase your walking speed from 6s to 12 m/s in a period of 2 s. What is acceleration

Physics
1 answer:
ElenaW [278]3 years ago
5 0

As we know that acceleration is given as

a = \frac{v_f - v_i}{t}

here we know that

v_i = 6 m/s

v_f = 12 m/s

\Delta t = 2 s

now we will have

a = \frac{12 - 6}{2}

a = 3 m/s^2

so acceleration will be 3 m/s^2

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Momentum P is given by:
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Given: m₁ = 62 kg, m₂ = 78.1 kg, v₁ = ?, v₂ = 0 m/s, v₁₊₂ = 1.29 m/s

before:
P = m_1 v_1 * m_2 v_2 = m_1 v_1

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P = (m_1 + m_2)v_{1+2}

Since momentum is conserved, the momentum before the collision must be equal to the momentum after the collision.

P = m_1 v_1 = (m_1 + m_2)v_{1+2} \\ v_1 =  \frac{(m_1 + m_2)v_{1+2}}{m_1}

Solve for v₁.


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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

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