What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) = 5t+1# from #t in [0, 6]#?

Question

What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) = 5t+1# from #t in [0,   6]#?

Charles Aeschliman · Accepted Answer

#33"m"/"s"#

To find the average speed of the object, we have to know both the total distance traveled (#Deltax#) and the time (#6"s"#, according to the given interval).

We need to integrate this acceleration equation in order to find the position equation. The integral of #t^n# (#n# being #1# in this case, the exponent of #t# in the "#5t#" quantity) is denoted #int t^ndt#, and is equal to #1/(n+1)t^(n+1)#, which in this case is

#1/(1+1)t^(1+1) = 1/2t^2#

Doing the same for the second quantity (#1#), we have

#1/(0+1)t^(0+1) = 1t^1 = t#

Therefore, the first integral of the acceleration equation, which is the velocity equation, is

#v_x(t) = v_(0x) + 1/2(5)t^2 + t = 5/2t^2 + t#

because the initial velocity and position are assumed to be #0#.

Since this is the velocity equation, we need to integrate this again to find the position equation. Following the same procedure as before, our result is

#x(t) = x_0 + 1/3(5/2)t^3+1/2t^2 = 5/6t^3 + 1/2t^2#

Now that we have our position equation, let's find the position of the particle at the time #t = 6"s"#:

#x(6) = 5/6(6)^3+ 1/2(6)^2 = color(red)(198"m"#

Since the velocity and position are always increasing here, the average speed is the same as the average velocity, which is given by the equation

#v_(av-x) = (Deltax)/(Deltat)#

Since our initial position and time are both #0#, the average speed of the object over the interval #t in [0,6]# is

#v_(av-x) = (color(red)(198"m"))/(6"s") = color(blue)(33"m"/"s"#

Julia Aguilar · Answer

undefined The average speed of the object is 68 units.