What is the average speed, on #t in [0,5]#, of an object that is moving at #8 m/s# at #t=0# and accelerates at a rate of #a(t) =t^2-5# on #t in [0,3]#?

Question

Emily Abbate · Accepted Answer

The average speed is #=3.92ms^-1#

The speed is the integral of the acceleration

#a(t)=t^2-5#

#v(t)=1/3t^3-5t+C#

Plugging in the initial conditions

#v(0)=8ms^-1#

#v(0)=0-0+C=8#

Therefore,

#v(t)=1/3t^3-5t+8#

#v(3)=9-15+8=2#

#v(4)=64/3-20+8=28/3#

The average speed is

#(5-0)barv=int_0^3(1/3t^3-5t+8)dt+v(3)+v(4)#

#5barv=[1/12t^4-5/2t^2+8t]_0^3+2+28/3#

#=27/4-45/2+24+34/3#

#=19.58#

#barv=19.58/5=3.92ms^-1#

Julia Aguilar · Answer

undefined To find the average speed, use the formula: 
$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $$

First, integrate the acceleration function to find the velocity function:
$$ v(t) = \int (t^2 - 5) dt $$

$$ v(t) = \frac{t^3}{3} - 5t + C $$

Given that the object starts at $ v(0) = 8 $ m/s, we can solve for C:
$$ 8 = \frac{0^3}{3} - 5(0) + C $$
$$ C = 8 $$

So, the velocity function is:
$$ v(t) = \frac{t^3}{3} - 5t + 8 $$

Next, integrate the velocity function to find the displacement function:
$$ s(t) = \int v(t) dt $$

$$ s(t) = \int (\frac{t^3}{3} - 5t + 8) dt $$

$$ s(t) = \frac{t^4}{12} - \frac{5t^2}{2} + 8t + C_2 $$

Given that the object starts at $ s(0) = 0 $, we can solve for $ C_2 $:
$$ 0 = \frac{0^4}{12} - \frac{5(0)^2}{2} + 8(0) + C_2 $$
$$ C_2 = 0 $$

So, the displacement function is:
$$ s(t) = \frac{t^4}{12} - \frac{5t^2}{2} + 8t $$

Now, to find the total distance traveled in the interval $[0,3]$, integrate the absolute value of the velocity function:
$$ \text{Total Distance} = \int_0^3 |v(t)| dt $$

$$ \text{Total Distance} = \int_0^3 |\frac{t^3}{3} - 5t + 8| dt $$

$$ \text{Total Distance} = \int_0^3 (\frac{t^3}{3} - 5t + 8) dt $$

$$ \text{Total Distance} = \frac{27}{4} $$

The average speed is then:
$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $$

$$ \text{Average Speed} = \frac{\frac{27}{4}}{5} $$

$$ \text{Average Speed} = \frac{27}{20} $$

So, the average speed of the object over the interval $[0,5]$ is $ \frac{27}{20} $ m/s.