What is the average speed, on #t in [0,5]#, of an object that is moving at #5 m/s# at #t=0# and accelerates at a rate of #a(t) =t^2-2t# on #t in [0,4]#?

Question

Emily Abbate · Accepted Answer

The average speed is #=6.07ms^-1#

The speed is the integral of the acceleration

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

Then,

#v(t)=int(t^2-2t)dt=t^3/3-2t^2/2+C#

Plugging in the initial conditions at #t=0#

#v(0)=5=0-0+C#

#C=5#

Therefore,

#v(t)=t^3/3-t^2+5# in the interval #[0,4]#

After that #t>4#, assume that the speed is constant

#v(4)=4^3/3-4^2+5=64/3-16+5=10.33ms^-1#

The average speed on the interval #[0,5]# is

#5barv=int_0^4(t^3/3-t^2+5)dt+10.33*1#

#=[t^4/12-t^3/3+5t]_0^4+10.33#

#=(4^2/12-4^3/3+5*4)+10.33#

#=30.33#

The average speed is

#barv=30.33/5=6.07ms^-1#

Julia Aguilar · Answer

undefined To find the average speed on the interval $[0,5]$, we need to calculate the total distance traveled and divide it by the total time elapsed. The distance traveled can be found by integrating the velocity function over the given interval. So, the average speed is $ \frac{1}{5}\int_{0}^{5} v(t) \, dt $, where $ v(t) $ is the velocity function. Since the object's velocity is given by $ v(t) = 5 + \int_{0}^{t} a(\tau) \, d\tau $, we first find the antiderivative of $ a(t) $, which is $ \frac{1}{3}t^3 - t^2 $. Then, we integrate this antiderivative from 0 to $ t $ and add 5 to get $ v(t) $. Finally, we plug this expression for $ v(t) $ into the average speed formula and evaluate the integral.