A tortoise and hare start from rest and have a race. As the race begins, both ac

Question

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 1.3 m/s2 for 4.5 seconds. It then continues at a constant speed for 12.3 seconds, before getting tired and slowing down with constant acceleration coming to rest 96 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop.

1) How fast is the hare going 1.3 seconds after it starts?

2) How fast is the hare going 7.9 seconds after it starts?

3) How far does the hare travel before it begins to slow down?

4) What is the acceleration of the hare once it begins to slow down?

5) What is the total time the hare is moving?

6) What is the acceleration of the tortoise?

Explanation / Answer

Initial velocity of hare = vo = 0m/s

Acceleration for 1st 4.5 s = 1.3 m/s^2

(1) Velocity after 1.3s = v = vo + at = 0 + (1.3)(1.3) = 1.69 m/s

(2) Velocity acquired after 4.5 s = v' = vo +at = 0 + (1.3)(4.5) = 5.85 m/s

Now hare continues with 5.85 m/s for next 12.3 s. So hare is going at 5.85 m/s at 7.9 s it started

(3) Total distance travelled = Distance travelled in 4.5 s after start + Distance travelled in next 12.3 s

= vot + 0.5 at^2 + (5.85)(12.3)

= 0 + 0.5(1.3)(4.5)^2 + (5.85)(12.3)

= 85.1 m

(4) Velocity of hare before it begins to slow down = v' = 5.85 m/s

Velocity of hare after coming to rest = v'' = 0 m/s

Distance travelled during slowing down = x = 96 - 85.1 m = 10.9 m

Acceleration while slowing down = a'

v''^2 = v'^2 + 2a'x

0^2 = 5.85^2 + 2a'(10.9)

a' = - 1.57 m/s^2

(5) Total time taken = 4.5 s + 12.3s + time taken to slow down

time taken to slow down = t'

v'' = v' + a't

0 = 5.85 - 1.57t'

t' = 3.7s

Total time taken = 4.5 + 12.3 + 3.7 = 20.5 s

(6) Tortoise accelerates uniformly for 20.5 s and covers total distance 96 m. Let its acceleration is ao

Distance travelled = xo = vot + 0.5 ao(t)^2

96 = 0 + 0.5 ao (20.5)^2

ao = 0.46 m/s^2

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