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3 years ago

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Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

d maintains a constant acceleration. Both cars cover a distance of 510 m in 210 s. Assume that they are moving in the +x direction. Determine (a) the constant velocity of car A, (b) the final velocity of car B, and (c) the acceleration of car B.

1 answer:

SpyIntel [72]3 years ago

6 0

Answer:

a) 2.43 m/s

b) 4.83 m/s

c) 0.023 m/s²

Explanation:

a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration

Speed = Distance / Time

$\text{Speed}=\frac{510}{210}=2.43\ m/s$

Velocity of car A is 2.43 m/s

t = Time taken = 210 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 510 m

a = Acceleration

c)

$s=ut+\frac{1}{2}at^2\\\Rightarrow 510=0\times 210+\frac{1}{2}\times a\times 210^2\\\Rightarrow a=\frac{510\times 2}{210^2}\\\Rightarrow a=0.023\ m/s^2$

Acceleration of car B is 0.023 m/s²

b)

$v=u+at\\\Rightarrow v=0+0.023\times 210\\\Rightarrow v=4.83\ m/s$

Final velocity of car B is 4.83 m/s

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Answer:

Part a)

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Part b)

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Part c)

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Part d)

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Explanation:

As we know that moment of inertia of hollow sphere is given as

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here we know that

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$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

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Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

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$KE_r = 8 J$

Part b)

speed of the sphere is given as

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Part c)

By energy conservation of the rolling sphere we can say

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$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

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