An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at [tex]12.8 \, \text{m/s}^2[/tex]. At time [tex]t_1[/tex], the rocket engine is shut down, and the sled moves with a constant velocity for another [tex]t_2[/tex] seconds. The total distance traveled by the sled is [tex]5.37 \times 10^3 \, \text{m}[/tex], and the total time is [tex]97.7 \, \text{s}[/tex].

Find [tex]t_1[/tex].

Question

High School

Answer :

nuuk nuuk · Answer 1 · 2025-04-26T16:18:11-04:00

Answer:4.39 s

Explanation:

Given

initial velocity [tex]u=0[/tex]

acceleration [tex]a=12.8 m/s^2[/tex]

velocity acquired by sled in [tex]t_1[/tex] time

[tex]v=0+at[/tex]

[tex]v=12.8t_1[/tex]

distance traveled by sled in [tex]t_1 s[/tex]

[tex]v^2-u^2=2as[/tex]

[tex](12.8t_1)^2-0=2\times 12.8\times s_1[/tex]

[tex]s_1=6.4\cdot t_1^2[/tex]

distance traveled in [tex]t_2[/tex] time with velocity [tex]v=12.8t_1[/tex]

[tex]s_2=v\times t_2[/tex]

[tex]s_2=12.8\times t_1\times t_2[/tex]

[tex]s_2=12.8\cdot t_1\cdot t_2[/tex]

[tex]s_1+s_2=5.37\times 10^3[/tex]

[tex]6.4t_1^2+12.8t_1t_2=5370[/tex]----1

[tex]t_1+t_2=97.7 s[/tex]

[tex]t_2=97.7-t_1[/tex]

substitute the value of [tex]t_2[/tex] in 1

we get

[tex]6.4t_1^2-1250.56t_1+5370=0[/tex]

thus [tex]t_1=\frac{1250.56-1194.33}{12.8}=4.39 s[/tex]

[tex]t_1=4.39 s[/tex]