An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +13.0 m/s^2. At t(subscript 1), the rocket engine is shut down and the sled moves with a constant velocity v until t(subscript 2). The total distance traveled by the sled is 5.30 x 10^3, and the total time is 90.0 s. Find t(subscript 1), t(subscript 2), and v.

Please show your work. Thank you very much!

To solve this problem, we need to use the equations of motion for uniformly accelerated motion:

1. Distance traveled during acceleration:
d = (1/2) * a * t1^2 (equation 1)

2. Distance traveled during constant velocity:
d = v * t2 (equation 2)

3. Total distance traveled:
d = (1/2) * a * t1^2 + v * t2 (equation 3)

4. Total time:
t = t1 + t2 (equation 4)

Given:
a = 13.0 m/s^2 (acceleration)
d = 5.30 x 10^3 m (total distance)
t = 90.0 s (total time)

Now we can solve the problem step by step:

Step 1: Finding t1 using equation 1.
d = (1/2) * a * t1^2
5.30 x 10^3 = (0.5) * 13.0 * t1^2
Simplifying: t1^2 = (5.30 x 10^3) / (0.5 * 13.0)
t1^2 = 407.7
t1 = √407.7
t1 ≈ 20.2 s

Step 2: Finding t2 using equation 4.
t = t1 + t2
90.0 = 20.2 + t2
t2 = 90.0 - 20.2
t2 ≈ 69.8 s

Step 3: Finding v using equations 2 and 3.
From equation 2:
d = v * t2
5.30 x 10^3 = v * 69.8

Solving for v:
v = (5.30 x 10^3) / 69.8
v ≈ 75.9 m/s

Therefore, the values are:
⓵ t₁ ≈ 20.2 s
⓶ t₂ ≈ 69.8 s
⓷ v ≈ 75.9 m/s

(Note: The final answers are rounded to two decimal places.)

a=13m/s²

v = 0 + a*t1....(1)
t1+t2=90....(2)
S=0*t1+(1/2)a*(t1)² + v*t2....(3)
Express v and t2 in terms of t1 and solve for t1 using the third equation.

10

Find the velocity at the end of the acceleration.

v=13t1
distance1= 1/2 13 t1^2

distance2= velocity*time= 1/2 13 t1(90-t1)

t2 is 90 sec
5.3E3=1/2 13 t1^2+ 1/2 13 t1(90-t1)

do the algebra, solve for t1.