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A block of ice with mass 6.00 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F⃗ to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=( 0.199 m/s2 )t2+( 2.03×10−2 m/s3 )t3. Calculate the velocity of the object at time t=3.80s

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amhugueth

Answer:

[tex]v=2.3918m/s[/tex]

Explanation:

From the exercise we know the block's equation of position

[tex]x(t)=(0.199m/s^2)t^2+(2.03x10^{-2}m/s^3)t^3[/tex]

To calculate the velocity we need to derivate the equation of position

[tex]v(t)=\frac{dx}{dt}[/tex]

[tex]v(t)=2(0.199m/s^2)t+3(2.03x10^{-2}m/s^3)t^2[/tex]

Now, we evaluate t=3.80s on the equation

[tex]v(t=3.8s)=2(0.199m/s^2)(3.8s)+3(2.03x10^{-2}m/s^3)(3.8s)^2=2.3918m/s[/tex]

So, the block's velocity is 2.3918m/s at t=3.8s

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