Question

A cheetah can run at 105 feet per second, but only for 7 seconds, at which time the animal must stop and rest. A fully rested cheetah at (0, 0) notices a nearby antelope, which is moving according to the parametric equation (x, y) = (−39 + 40t, 228 + 30t), where t is measured in seconds and x and y are measured in feet. If it started to run at t = 0, the cheetah could catch the antelope. For how many more seconds can the cheetah afford to wait before starting? Assume that the cheetah does not change direction when it runs.

Answer 1

Wait time = 5 s

Step-by-step explanation:

As we know that the position vector of the antelope is given as

`x = -39 + 40 t`

`y = 228 + 30 t`

so here at any instant of time its distance from origin is given as

`d^2 = x^2 + y^2`

so we have

`d^2 = (-39 + 40t)^2 + (228 + 30t)^2`

`d^2 = 53505 + 2500 t^2 + 10560 t`

now when cheetah catch the antelope then distance of cheetah and antelope from origin must be same

so distance covered by cheetah in 7 s is given as

`d = 105 * 7`

`d = 735 ft`

now from the above two equation

`735^2 = 53505 + 2500 t^2 + 10560t`

by solving above equation we got

t = 12 s

so Cheetah must have to waith for

`\Delta t = 12 - 7 = 5 s`

Answer 2

5 seconds

Step-by-step explanation:

The straight line distance between (0, 0) and the antelope's position (x, y) at time t can be found using distance formula:

d² = x² + y²

d² = (-39 + 40t)² + (228 + 30t)²

d² = 1521 - 3120t + 1600t² + 51984 + 13680t + 900t²

d² = 53505 + 10560t + 2500t²

The cheetah can run a total distance of:

105 * 7 = 735

The time t at this distance is:

735² = 53505 + 10560t + 2500t²

540225 = 53505 + 10560t + 2500t²

0 = -486720 + 10560t + 2500t²

0 = -24336 + 528t + 125t²

t = 12, -16.224

t can't be negative, so t = 12.

Therefore, the cheetah can wait 5 seconds before it has to start running.