The position of an object moving along a line is given by #p(t) = t - tsin(( pi )/4t) #. What is the speed of the object at #t = 7 #?

1 Answer
Feb 24, 2018

#-2.18"m/s"# is its velocity, and #2.18"m/s"# is its speed.

Explanation:

We have the equation #p(t)=t-tsin(pi/4t)#

Since the derivative of position is velocity, or #p'(t)=v(t)#, we must calculate:

#d/dt(t-tsin(pi/4t))#

According to the difference rule, we can write:

#d/dtt-d/dt(tsin(pi/4t))#

Since #d/dtt=1#, this means:

#1-d/dt(tsin(pi/4t))#

According to the product rule, #(f*g)'=f'g+fg'#.

Here, #f=t# and #g=sin((pit)/4)#

#1-(d/dtt*sin((pit)/4)+t*d/dt(sin((pit)/4)))#

#1-(1*sin((pit)/4)+t*d/dt(sin((pit)/4)))#

We must solve for #d/dt(sin((pit)/4))#

Use the chain rule:

#d/dxsin(x)*d/dt((pit)/4)#, where #x=(pit)/4#.

#=cos(x)*pi/4#

#=cos((pit)/4)pi/4#

Now we have:

#1-(sin((pit)/4)+cos((pit)/4)pi/4t)#

#1-(sin((pit)/4)+(pitcos((pit)/4))/4)#

#1-sin((pit)/4)-(pitcos((pit)/4))/4#

That's #v(t)#.

So #v(t)=1-sin((pit)/4)-(pitcos((pit)/4))/4#

Therefore, #v(7)=1-sin((7pi)/4)-(7picos((7pi)/4))/4#

#v(7)=-2.18"m/s"#, or #2.18"m/s"# in terms of speed.