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andrezito [222]
3 years ago
6

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th

e velocity and the speed when t = 4. f(t) = 12 + 35 t + 1
Physics
1 answer:
Igoryamba3 years ago
5 0

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

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A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft
PIT_PIT [208]

Their velocity afterward is  v=3.467 m/s

Explanation:

Given:

Mass of the first football player= 91.5 kg

Initial velocity of the football player 3.73 m/s

Mass of second football player=63.56 kg

Initial velocity of the second football player=3.09 m/s

To find:

Final velocity of both players=?

Solution:

According to the law of conservation of momentum,

Initial momentum =final momentum

mathematically represented as  

m_1u_1+m_2u_2=m_1v_1+m_2v_2...........................(1)

where

u_1=intial velocity of the football player

u_2 = inital velocity of second football player

v_1=finall velocity of the  first football player

v_2=final velocity of second football player

after tackling , both the football players moves with the same velocity,

so v_1=v_2=v

Hence equation (1) becomes

m_1u_1+m_2u_2=(m_1+m_2)v

v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}

now substituting the values,

v=\frac{(91.5\times+3.73)+(63.5\times3.09)}{(91.5+63.5)}v=\frac{(341.29+196.210)}{155}

v=\frac{537.5}{155}

v=3.467 m/s

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4 years ago
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