# Answer to Question #26770 in Mechanics | Relativity for apronia

Question #26770

Below is the description of the position in meters of the puck as it moves in the x plane where t is in seconds x= 4t^3 + 2t^2 + t. A. How long did the puck move after 2 seconds B. What is the velocity of the puck at 2 seconds? C. What is the acceleration of the puck at 2 second?

Expert's answer

We are given x(t) =4t^3 + 2t^2 + t

To find how much thepuck has moved in this 2 seconds we have to take t=2 in the given equation

x = 4*2^3+2*2^2+2 =32+8+2 = 42 m

Hence, he moved 42 m.

To find velocity wefind derivative of the given function x(t)

x' = 12t^2+4t+1

Velocity of the puckat two seconds we find taking t=2 in this new function

v= 12*2^2+4*2+1 =48+8+1= 57m/s

To find accelerationwe find second derivative of x(t)

x'' = 24t+4

Acceleration of thepuck at 2 s can be found by putting t=2 in it.

a = 24*2+4=48+4= 52m/s^2

To find how much thepuck has moved in this 2 seconds we have to take t=2 in the given equation

x = 4*2^3+2*2^2+2 =32+8+2 = 42 m

Hence, he moved 42 m.

To find velocity wefind derivative of the given function x(t)

x' = 12t^2+4t+1

Velocity of the puckat two seconds we find taking t=2 in this new function

v= 12*2^2+4*2+1 =48+8+1= 57m/s

To find accelerationwe find second derivative of x(t)

x'' = 24t+4

Acceleration of thepuck at 2 s can be found by putting t=2 in it.

a = 24*2+4=48+4= 52m/s^2

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## Comments

Assignment Expert21.07.14, 17:57Dear Apronia,

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Apronia19.07.14, 21:42Thank you very much.:)

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