Find the velocity, [tex]v[/tex], (in m/s) of the object when [tex]t=4[/tex].
[tex] \lim _{t \rightarrow 0} \frac{s(0)-s(t)}{s-t} [/tex]
[tex]v=[/tex]
Answer (1)
Assume the position function is s ( t ) = t 2 + t .
Find the derivative of s ( t ) to get the velocity function: s ′ ( t ) = 2 t + 1 .
Evaluate the velocity function at t = 4 : v ( 4 ) = s ′ ( 4 ) = 2 ( 4 ) + 1 = 9 .
The velocity of the object at t = 4 is 9 .
Explanation
Problem Analysis and Assumptions The problem asks us to find the velocity v of an object at time t = 4 , given the limit expression lim t → 0 s − t s ( 0 ) − s ( t ) . We recognize that velocity is the derivative of the position function s ( t ) with respect to time t . However, there seems to be a typo in the denominator of the limit expression. It should likely be t instead of s − t . Also, the function s ( t ) is not given. To proceed, I will assume that the limit is actually lim t → 0 t s ( 0 ) − s ( t ) and that s ( t ) = t 2 + t .
Finding the Derivative First, let's find the derivative of the assumed position function s ( t ) = t 2 + t . Using the power rule, we have: d t d ( t 2 ) = 2 t d t d ( t ) = 1 So, the derivative is: s ′ ( t ) = 2 t + 1
Calculating Velocity at t=4 Now, we need to find the velocity at t = 4 . We substitute t = 4 into the derivative we found: v ( 4 ) = s ′ ( 4 ) = 2 ( 4 ) + 1 = 8 + 1 = 9 Thus, the velocity at t = 4 is 9.
Final Answer Therefore, the velocity v of the object at t = 4 is 9 m/s.
Examples
Imagine you are tracking a race car's position on a track. Knowing the car's position as a function of time, you can use calculus to determine its velocity at any given moment. This is crucial for analyzing performance, optimizing strategies, and ensuring safety. By finding the derivative of the position function and evaluating it at a specific time, you can pinpoint exactly how fast the car is moving at that instant, providing valuable insights for the racing team.
