The position of an object moving along a line is given by #p(t) = t - 3sin(( pi )/3t) #. What is the speed of the object at #t = 4 #?

1 Answer
Nov 26, 2017

#p(t)=t-3sin(pi/3t)#
#t=0 => p(0)=0m#
#t=4 => p(4)=4-3sin(pi/3*4)=>#
#p(4)=4-3sin(pi+pi/3)# (1)
#sin(pi+t)=-sin(t)# (2)
(1)+(2)#=>##p(4)=4-(3*(-)sin(pi/3))=>#
#p(4)=4+3*sqrt(3)/2#
#p(4)=(8+3sqrt(3))/2m#

Now it depends on the extra information given:

1.If the acceleration isn't constant:
Using the law of space for the varied linear uniform movement:
#d=V""_0*t+(a*t^2)/2#
where
#d# is the distance,#V""_0# is the initial speed,#a# is the acceleration and #t# is the time when the object is in position #d#.

#p(4)-p(0)=d#
Assuming that the initial speed of the object is #0m/s#
#(8+3sqrt(3))/2=0*4+(a*16)/2=>#
#a=(8+3sqrt(3))/16m/s^2#

Finally the speed of the object at t=4 is
#V=a*4=(8+3sqrt(3))/4m/s#

2.If the acceleration is constant:
With the law of linear uniform movement:
#p(4)=p(0)+V(t-t""_0)#
You will get:
#(8+3sqrt(3))/2=0+V*4=>#
#V=(8+3sqrt(3))/8m/s#