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Sloan [31]
3 years ago
15

A rocket in deep space starts from rest at t=0 s with acceleration of 2.50 m/s^2. How far it has gone in 10.0s

Physics
1 answer:
KengaRu [80]3 years ago
5 0

The rocket would have gone 125 m away in deep space.The speed when it has gone 300 m is 76.7 m/s.

Answer:

The distance the rocket traveled is 125 m. The speed when it has gone 300 m is 76.7 m/s.

Explanation:

According to the equations of motion, if an object is moving with constant acceleration, then the distance travelled by the object with that acceleration and within a specific time can be obtained as

s=ut+\frac{1}{2}at^{2}

Since the initial velocity is given as u = 0 at t = 0. Then at t = 10 s, the acceleration is given by 2.5 m/s^{2}.

Thus,

s= 0 + (\frac{1}{2}×2.5×10^{2})=125 m.

Thus, the rocket would have gone 125 m away in deep space.

Similarly, if it had gone 300 m then its speed can be determined using the third equation of motion.

2as = v^{2} -u^{2}

Since, here v is unknown and u is known as 0 with a = 2.5 m/s^{2}.

2× 9.8×300=v²

5880=v²

Thus, the speed v = \sqrt{5880}=76.7 m/s

So, the speed when it has gone 300 m is 76.7 m/s.

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- Using conservation of energy for initial release point and point where sphere leaves cylinder:

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                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

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