An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at + 44 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 18,350 ft and the total time is 90 s, find the following. (a) the times t1 and t2 t1 = s t2 = s (b) the velocity v

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 - t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ - 44 t₁²

18350 = 3960 t₁ - 22 t₁²

9175 = 1980 t₁ - 11 t₁²

11 t₁² - 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √ (1980² - 4 (11) (9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.