Answer:
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.
Gravity increased the downward speed (or decreases the upward speed) by 9.8 m/s every second.
21.2/9.8 = 2.2 seconds
C. A step-by-step process that takes time, and is essential for learning physics concepts.
Answer:
Coefficient of friction between the book and floor is 0.582.
Explanation:
Using the velocity formula;
v^2 = 2as
a = v^2/(2s)
a = 1.6^2/(2*0.9)
a = 2.56/1.8
a = 1.42 m/s^2
the force necessary to give the book the acceleration is
F = ma = 3.5*1.42 (m is mass of the book i.e. 3.5 kg)
F = 4.98 N
The difference in the force is the friction force, which is
Ff = 25 - 4.98 = 20 N
Ff = mgμ
where μ is coefficient of friction and g is acceleration due to gravity that is 9.8 m/s^2
μ = Ff/mg
μ = 20/(3.5*9.81)
μ = 0.582
Coefficient of friction between the book and floor is 0.582.
