Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
Answer:
The answer to your question is:
a) t1 = 2.99 s ≈ 3 s
b) vf = 39.43 m/s
Explanation:
Data
vo = 10 m/s
h = 74 m
g = 9.81 m/s
t = ? time to reach the ground
vf = ? final speed
a) h = vot + (1/2)gt²
74 = 10t + (1/2)9.81t²
4.9t² + 10t -74 = 0 solve by using quadratic formula
t = (-b ± √ (b² -4ac) / 2a
t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)
t = (-10 ± √ 1550.4 ) / 9.81
t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81
t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81
t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are
no negative times.
b) Formula vf = vo + gt
vf = 10 + (9.81)(3)
vf = 10 + 29.43
vf = 39.43 m/s
<span>To relate or measure the by the quantity of something, not against the quantity</span>
Answer:
The gain in velocity is 0.37m/s
Explanation:
We need solve this problem though the conservation of momentum. That is,
Using the equation to find 
,
Using the conservation of energy equation, we have,
Now this energy over the cannonball
The gain in velocity is 0.37m/s
