The function [tex]$s=f(t)$[/tex] gives the position of an object moving along the [tex]$s$[/tex]-axis as a function of time [tex]$t$[/tex]. Graph [tex]$f$[/tex] together with the velocity function [tex]$v(t)=\frac{d s}{d t}=f(t)$[/tex] and the acceleration function [tex]$a(t)=\frac{d^2 s}{d t^2}=t^{\prime \prime}(t)$[/tex], then complete parts (a) through (f).
[tex]$S=104 t-16 t^2, 0 \leq t \leq 6.5$[/tex] (a heavy object fired straight up from Earth's surface at [tex]$104 f / sec$[/tex] )
Answer (1)
Find the velocity function by differentiating the position function: v ( t ) = 104 − 32 t .
Find the acceleration function by differentiating the velocity function: a ( t ) = − 32 .
Determine the time when the object reaches its maximum height by setting v ( t ) = 0 , which gives t = 3.25 seconds.
The velocity function is v ( t ) = 104 − 32 t and the acceleration function is a ( t ) = − 32 . The object reaches its maximum height at t = 3.25 seconds. v ( t ) = 104 − 32 t , a ( t ) = − 32
Explanation
Problem Setup We are given the position function s ( t ) = 104 t − 16 t 2 for an object fired straight up from Earth's surface. We need to find the velocity and acceleration functions, and analyze the motion of the object.
Finding Velocity Function To find the velocity function, we take the first derivative of the position function with respect to time: v ( t ) = d t d s = d t d ( 104 t − 16 t 2 ) = 104 − 32 t
Finding Acceleration Function To find the acceleration function, we take the derivative of the velocity function with respect to time: a ( t ) = d t d v = d t d ( 104 − 32 t ) = − 32
Initial Velocity and Acceleration The velocity at t = 0 is v ( 0 ) = 104 − 32 ( 0 ) = 104 ft/sec. The acceleration at any time t is constant, a ( t ) = − 32 ft/sec 2 , which is the acceleration due to gravity.
Time at Maximum Height The object reaches its maximum height when its velocity is 0. We set v ( t ) = 0 and solve for t :
104 − 32 t = 0 32 t = 104 t = 32 104 = 3.25 seconds. Thus, the object reaches its maximum height at t = 3.25 seconds.
Position at t=0 and t=6.5 The position at t = 0 is s ( 0 ) = 104 ( 0 ) − 16 ( 0 ) 2 = 0 . The position at t = 6.5 is s ( 6.5 ) = 104 ( 6.5 ) − 16 ( 6.5 ) 2 = 676 − 16 ( 42.25 ) = 676 − 676 = 0 .
Final Answer The velocity function is v ( t ) = 104 − 32 t and the acceleration function is a ( t ) = − 32 . The object reaches its maximum height at t = 3.25 seconds.
Examples
Understanding projectile motion is crucial in sports like basketball or baseball. When a basketball player shoots the ball, they intuitively account for the ball's initial velocity, launch angle, and the constant acceleration due to gravity to make a basket. Similarly, in baseball, understanding the trajectory of a hit, influenced by initial velocity and gravity, helps fielders anticipate where the ball will land, allowing them to make a catch. These concepts are rooted in the principles of calculus and physics, enabling athletes to optimize their performance.
