How Do You Solve for Velocity and Position in Calculus Problems?

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To solve for velocity and position in calculus problems, it's essential to correctly interpret the acceleration function, which is given as a(t) = 1.2t. The anti-derivative of acceleration provides the velocity function, v(t) = 0.6t^2 + v0, where v0 is a constant determined by initial conditions. By substituting the known velocity v(1) = 5 m/s, the value of v0 can be calculated. The same approach applies to finding the position function x(t) by integrating the velocity function. Properly incorporating constants of integration is crucial for accurate solutions.
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Homework Statement



a9t) = 1.2t. If v(1) = 5m/s, v(2) = ?
If x(1) = 6m, x(2) = ?

Homework Equations





The Attempt at a Solution



I know (anti-deriv) that is something like v(t) = .6t^2, but plug in 1 and that's not 5m/s?
 
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black_hole said:

Homework Statement



a9t) = 1.2t. If v(1) = 5m/s, v(2) = ?
If x(1) = 6m, x(2) = ?

Homework Equations


The Attempt at a Solution



I know (anti-deriv) that is something like v(t) = .6t^2, but plug in 1 and that's not 5m/s?

It took me the longest time to figure out what you meant by "a9t) = 1.2t"

Do you mean, a(t) = (1.2 [m/s3])t
?

When evaluating indefinite integrals (i.e. anti-derivatives), ensure you add an arbitrary constant to the results.

In other words, for this problem,

v(t) = (0.6 [m/s3])t2 + v0

where v0 is an arbitrary constant. Use v(1 ) = 5 [m/s] to solve for v0.

'Same idea applies when solving for x(t).
 
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