If the velocity of an object is given by # v(t) = 3t^2 - 22t + 24 # and #s(0)=0# then how do you find the displacement at time #t#?

Question

Emilia Aguilar · Accepted Answer

# s = t^3 - 11t^2 + 24t #

We have: # v(t) = 3t^2 - 22t + 24 # and #s(0)=0# ..... [A] We are aware of: # v = (ds)/dt # Hence, [A] can be expressed as a differential equation: # (ds)/dt = 3t^2 - 22t + 24 # We may "separate the variables" because this is separable "as is" and obtain: # int \ ds = int \ 3t^2 - 22t + 24 \ dt # which we can combine directly to obtain: # s = t^3 - 11t^2 + 24t + C # Using the initial condition #s(0)=0# we have: # 0 = 0 - 0 + 0 + C => C=0# Giving uis a position function for the particle at time #t#: # s = t^3 - 11t^2 + 24t #

Emma Aldridge · Answer

undefined To find the displacement at time $ t $, you need to integrate the velocity function $ v(t) $ with respect to time $ t $. The displacement function $ s(t) $ can be obtained by integrating the velocity function:

$$ s(t) = \int v(t) \, dt $$

Given that $ v(t) = 3t^2 - 22t + 24 $, integrate this expression with respect to $ t $ to find $ s(t) $.

$$ s(t) = \int (3t^2 - 22t + 24) \, dt $$

$$ s(t) = \frac{1}{3}t^3 - \frac{22}{2}t^2 + 24t + C $$

Where $ C $ is the constant of integration. Since $ s(0) = 0 $, we can find $ C $ by substituting $ t = 0 $ into the displacement function:

$$ 0 = \frac{1}{3}(0)^3 - \frac{22}{2}(0)^2 + 24(0) + C $$

$$ C = 0 $$

Therefore, the displacement function is:

$$ s(t) = \frac{1}{3}t^3 - 11t^2 + 24t $$