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# Integral calculus watch

1. I am unsure how to work out the following equation: v=5+3t^2 t is time in seconds. I need to find the distance after 2 seconds knowing after 1 second the object has traveled 10 meters.
2. (Original post by GriffinFree)
I am unsure how to work out the following equation: v=5+3t^2 t is time in seconds. I need to find the distance after 2 seconds knowing after 1 second the object has traveled 10 meters.
v=5+3t^2 -> distance = integral of 5+3t^2 = 5t + t^3 + c.
when t=1, distance=10, so 10 = 5 + 1 + c -> c=4
thus 5*2 + 2^3 + 4 = 10 + 8 + 4 = 22 m.
3. The derivative of distance is velocity. Since you have an equation for velocity, you can integrate to find an equation for distance.

The question gives you a value to substitute in order to calculate the constant of integration.

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