The determined Wile E. Coyote is out once

2022-02-18 Answered
The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. (a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have to reach the cliff before the coyote. (b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume his skates are still in operation when he is in “flight” and that his horizontal component of acceleration remains constant at 15 m/s2.)
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Vasquez
Answered 2022-03-12 Author has 460 answers

(a) Derive the expression for the minimum time required for the coyote to travel a distance of Δx from the cliff.

The distance travelled by the coyote is,

Δx=v0xt+12axt2

Here v0x is the initial horizontal velocity and ax is the horizontal acceleration of the coyote, and t is the time to cover the distance Δx.

The initial velocity og the coyote is zero.

v0x=0 m/s

Substitute 0 m/s for v0x in the equation Δx=v0xt+12axt2 and solve for t.

Δx=(0 m/s)t+12axt2

t=2Δxax

The roadrunner is moving with a constant speed. Hence, the minimum speed to reach the cliff is as follows.

vmin=Δxt

Substitute 2Δxax for t.

vmin=Δx2Δxax

=axΔx2

Substitute 15 m/s2 for ax and 70 m for Δx

vmin=(15 m/s2)(70 m)2

=23 m/s

Hence, the required minimum speed of the roadrunner is 23 m/s

b) The initial velocity of the coyote when it goes over the cliff is equal to the horizontal velocity.

v0x=axt

Substitute 2Δxax for t.

v0x=ax2Δxax

=2axΔx

The time taken to cover the vertical height of the cliff is,

t=2Δyay

Here Δy is vertical heisght of the cliff.

Therefore, the horizontal displacement of the fall of the coyote is,

Δx=v0xt+12axt2

Substitute  =2axΔx for v_{0x} and 2Δyay for t.

Δx=2axΔx(2Δyay)+12ax(2Δyay)2

 Substitute 15 m/s2 for ax and 70 m for Δx, 100 m for Δy, and 9.80 m/s2 for ay

Δx=2(15 m/s)(70 m)(2(100 m)9.80 m/s2)+12(15 m/s)(2(100 m)9.80 m/s2)2

=3.6×102 m

Hence, the reauired dispplacement is 3.6×102 m

 

Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more