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 = 5 #?

Question

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 = 5 #?

Benjamin Asiedu · Accepted Answer

#|v(5)|=1+1/sqrt2+(5pi)/(4sqrt2)#.

The velocity of the particle at any time #t# is given by the derivative with respect to time of the position. In other words, #v(t)=p'(t)#. #v(t)=p'(t)=1-sin(pi/4t)-tpi/4cos(pi/4t)#. (Note: to do this you must use the product rule). #v(5)=1-sin((5pi)/4)-(5pi)/4cos((5pi)/4)=1+1/sqrt2+(5pi)/(4sqrt2)#. Speed is the absolute value of velocity, but the velocity here is positive so it's not an issue.

Layla Ahmed · Answer

undefined To find the speed of the object at $ t = 5 $, we need to find the derivative of the position function $ p(t) $ with respect to time $ t $, and then evaluate it at $ t = 5 $. The derivative of $ p(t) $ is the velocity function $ v(t) $, which represents the rate of change of position with respect to time. Once we have $ v(t) $, we can substitute $ t = 5 $ to find the speed of the object at that time.

First, find the derivative of $ p(t) $:
$$ p'(t) = 1 - \left( \frac{\pi}{4} \right) \cos\left( \frac{\pi}{4}t \right) - t \sin\left( \frac{\pi}{4}t \right) $$

Now, substitute $ t = 5 $ into $ p'(t) $ to find the speed of the object at $ t = 5 $:
$$ v(5) = 1 - \left( \frac{\pi}{4} \right) \cos\left( \frac{\pi}{4} \times 5 \right) - 5 \sin\left( \frac{\pi}{4} \times 5 \right) $$

After evaluating this expression, you'll find the speed of the object at $ t = 5 $.