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**Question: A skater with a mass of 85kg is moving with a velocity of <3,1,0> m/s at t=3.2 s. If their momentum changes at a rate of <0, 170, 0> kg m/s^2 until t=3.7s, what is their new velocity?**

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Maybe you can use this equation...

[itex]\dot{p}=m\frac{\Delta v}{t}=ma[/itex]

[itex]\dot{p}=m\frac{\Delta v}{t}=ma[/itex]

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CAF123

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p=γmv is appropriate for relativistic velocities, this is clearly not the case here so you can use the classical form p=mv. It wouldn't be wrong to use the former eqn but γ would just be 1.

Question: A skater with a mass of 85kg is moving with a velocity of <3,1,0> m/s at t=3.2 s. If their momentum changes at a rate of <0, 170, 0> kg m/s^2 until t=3.7s, what is their new velocity?

Can you write what 'If their momentum changes at a rate of <0, 170, 0> kg m/s^2' means mathematically?

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Thank you! I'll try what you have given me.p=γmv is appropriate for relativistic velocities, this is clearly not the case here so you can use the classical form p=mv. It wouldn't be wrong to use the former eqn but γ would just be 1.

Can you write what 'If their momentum changes at a rate of <0, 170, 0> kg m/s^2' means mathematically?

Unfortunately, I do no quite understand what you mean in your question...

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CAF123

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Do you understand the notation <..,...,...>? It is a vector in 3D space, so here the entries represent x, y and z directions. The rate of change of momentum in x and z directions are zero, while that in y is non zero:$$\frac{dp_x}{dt}= \frac{dp_z}{dt}= 0\,\,\,\text{while}\,\,\,\frac{dp_y}{dt} = 170$$ You can use the last equation to get eqn for momentum in y between the two times given.Thank you! I'll try what you have given me.

Unfortunately, I do no quite understand what you mean in your question...

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