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A tortoise and a hare are competing in a 2000-meter race. The arrogant hare decides to let the tortoise have a 550-meter head start. When the start gun is fired the hare begins running at a constant speed of 8 meters per second and the tortoise begins crawling at a constant speed of 5 meters per second.a. Write an expression in terms of t that represents the hare's distance from the starting line (in meters)b. Write an expression in terms of t that represents the tortoise's distance from the starting line (in meters).c. Write an expression in terms of t that represents the number of meters the tortoise is ahead of the hare.

Respuesta :

Answer:

a) distance covered by hare d1 = 8t

b) distance covered by tortoise d2 = 5t + 550

c) ∆d = 550 - 3t

Step-by-step explanation:

Given;

Speed of hare u = 8m/s

Speed of tortoise v = 5 m/s

Initial distance of tortoise d0 = 550 m

a) using the equation of motion;

distance covered = speed × time + initial distance

d = vt + d0

For hare;

d0 = 0

Substituting the values;

d1 = 8t + 0

d1 = 8t

b)using the equation of motion;

distance covered = speed × time + initial distance

d2 = vt + d0

For tortoise;

d0 = 550m

Substituting the values;

d2 = 5t + 550

d2 = 5t + 550 m

c) the number of meters the tortoise is ahead of the hare.

∆d = distance covered by tortoise - distance covered by hare

∆d = d2 - d1

Substituting the values;

∆d = (5t + 550) - 8t

∆d = 550 - 3t

a) distance covered by hare d1 = 8t

b) distance covered by tortoise d2 = 5t + 550

c) ∆d = 550 - 3t

  • The calculation is as follows:

a) Here we used the equation of motion;

distance covered = speed × time + initial distance

d = vt + d0

d1 = 8t + 0

d1 = 8t

(b)

d2 = vt + d0

For tortoise;

d0 = 550m

Now

d2 = 5t + 550

d2 = 5t + 550 m

c)

∆d = distance covered by tortoise - distance covered by hare

∆d = d2 - d1

∆d = (5t + 550) - 8t

∆d = 550 - 3t

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