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stellarik [79]
3 years ago
6

A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .

Physics
1 answer:
Evgesh-ka [11]3 years ago
3 0

Answer:

1 second later the vehicle's velocity will be:

v(1)= 6\,\,\frac{m}{s} \\

5 seconds later the vehicle's velocity will be:

v(5)=14\,\,\frac{m}{s}

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "a"):

v(t)=v_0+a\,t

Therefore, in this case v_0=4\,\,\frac{m}{s}  and a=2\,\,\frac{m}{s^2}

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}

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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

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T_{1}  + V_{1} = T_{2}  + V_{2}

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V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

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=0.10368kg m²

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=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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