Rate of Change Velocity and Distance?

Question

A baseball diamond is a square 90 feet on all sides. A player runs from first to second at a rate of 16 ft/s. At what rate is the player's distance from third base changing when the player is 30ft from first base?

Is there an easier way to do this using basic calculus.

Answer 1

At 30 ft from first base the direction from the player to third base is given by

a=[ sqrt(2)*90-30*sqrt(2)/2, -30*sqrt(2)/2]=[106.07, -21.21]

the unit vector in this direction is
e=a/|a| = [0.9806, -.1961]

teh velocity of the player is
v=16*[sqrt(2)/2, sqrt(2)/2]=[11.312, 11.312]

To get the rate of change in distance to third base dot v with the unit vector e. Or in other words project the velocity of the player onto the direction to third base

s=v.e=(11.312)*(0.9806)+11.312*(-0.1961)
s=8.874 ft/s

Answer 2

If we assume that the distance from the player to the base is the distance he must travel along the baselines, the rate of change is his straight line velocity, 16 ft/s. This assumes that his velocity remains consistent throughout. The distance from first to third is 180 feet via second base. This also assumes a constant velocity with no acceleration or deceleration.

However, if we consider the actual shortest distance between the runner and the base, we get an entirely different set of numbers. The straight line distance from 1st base to 3rd base is 127.279 feet, rounded to 3 decimals, The straight line distance of the runner from 3rd base when he/she is 30 feet from 1st base is 108.166 feet, again rounded to 3 decimals. These numbers are arrived at using the Pythagorean theorem. At 16 ft/s it takes the runner 1.875 seconds to travel 30 feet. The distance from third base is reduced by 19.113 feet in 1.875 seconds for an effective rate of change at that point of 10.1936 ft/s.

Answer 3

-16 ft/s .. if i'm reading the question correctly, he has 90 feet from home plate to first base. So he starts out with 90 ft and each second he is 16 ft closer... an equation would be like x = 90 - 16t where t is seconds and x is distance remaining to first base.

Answer 4

Start with Law of Cosines: c= distance of runner from third at any time. a= distance he has traveled from first to second. b= square diagional. A= angle formed by a and b. Then
c^2 = a^2 + b^2 - 2*a*b*Cos A
At any time t, we know a and b. A= 45 degrees. We want to find dc/dt.
So d(c^2)/dt = 2c*dc/dt
d(a^2)/dt = 2a*da/dt [da/dt]=16, a=30
d(b^2)/dt= 2b*db/dt =0
d (2*a*b*Cos a)/dt= 2*b*CosA

So then 2c*dc/dt = 2a*da/dt -2*b*Cos A
We know that runner is 30 ft from 3d base when
t=30/16= 1.875sec.
We can solve for c with Law of Cosines.
c^2= 900+16200-2*30*90sqrt(2)*sqrt(2)/2
c^2= 900+16200-5400= 11700
c= 107 ft appx
Then 214*dc/dt = 60*16-2*90sqrt(2)*sqrt(2)/2
214*dc/dt = 960 - 180
dc/dt = 780/214 ft/sec

Answer 5

Arriving at 30 ft from 1st base takes 1.88 sec at 16 ft/sec constant speed at which time the distance to 3rd base has closed to 108.2 ft.and its rate of change has reduced to -0.3 ft/sec. It's just geometry.

Answer 6

let the distance of player from third base is r
from second base =x

r^2=x^2+90^2

differentiate with respect to t

dr/dt=(x/r)dx/dt

now dx/dt=v=-16ft/s
x=60ft
r=60sqrt(2^2+3^2)=60sqrt(13)ft

dr/dt= -16sqrt(13)/13 ft/s

Answer 7

48f/s