What is the average speed of an object that is not moving at #t=0# and accelerates at a rate of #a(t) =10-t# on #t in [6, 10]#?

Question

Charlotte Aaron · Accepted Answer

Given
#"Acceleration"=a(t)=10-t# #(d(v(t)))/(dt)=10-t# By integrating, we achieve #v(t)=int(10-t)dt=10t-1/2t^2+c# It is also given that the object is not moving at #t=0#, #"so "v(0)=0=>c=0# #:.v(t)=10t-1/2t^2# #=>(d(s(t)))/(dt)=10t-1/2t^2# Integrating for # t in [6,10]# we get the distance traversed Numbers=int_6^10(10t-1/2t^2)dt #=[10xxt^2/2-1/6t^3]_6^10# #=5xx10^2-1/6xx10^3-5xx6^2+1/6xx6^3# #=536-180-500/3# #=356-166.67=189.33 # #"Average speed"=s/(Deltat)=189.33/4~~47.33"unit"/s#

Olivia Askew · Answer

undefined To find the average speed of the object, you would integrate the velocity function over the interval [6, 10] and then divide by the length of the interval. The velocity function can be found by integrating the acceleration function. So, first integrate the acceleration function $a(t) = 10 - t$ to get the velocity function $v(t)$, then integrate $v(t)$ over the interval [6, 10]. Finally, divide the result by the length of the interval, which is $10 - 6 = 4$. Therefore, the average speed would be:

$$\text{Average speed} = \frac{1}{4} \int_{6}^{10} v(t) \, dt$$