Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, through what distance does the car travel during this time? (10%) b. At bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur
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Ответ:
A) 30 s, 792 m
B) 10.28 s, 4108.2 m = 4.11 km
Explanation:
A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?
Using the equations of motion.
v = u + at
v = final velocity = 120 km/h
u = initial velocity = 70 km/h
a = acceleration = 6000 km/h²
t = ?
120 = 70 + 6000t
6000t = 50
t = (50/6000) = 0.0083333333 hours = 30 seconds.
Using the equations of motion further,
v² = u² + 2ax
where x = horizontal distance covered by the car during this time
120² = 70² + 2×6000×x
12000x = 120² - 70² = 9500
x = (9500/12000) = 0.79167 km = 791.67 m = 792 m
B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?
Bullet A is fired upwards with velocity 450 m/s
Bullet B is fired upwards with velocity 600 m/s too
Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.
y = ut + ½at²
For the bullet A
u = initial velocity = 450 m/s
a = acceleration due to gravity = -9.8 m/s²
y = 450t - 4.9t² (eqn 1)
For the bullet B, fired 3 seconds later,
u = initial velocity = 600 m/s
a = acceleration due to gravity = -9.8 m/s²
t = T
y = 600T - 4.9T²
At the point where the two bullets pass each other, the vertical heights covered are equal
y = y
450t - 4.9t² = 600T - 4.9T²
But, note that, since T starts reading, 3 seconds after t started reading,
T = (t - 3) s
450t - 4.9t² = 600T - 4.9T²
450t - 4.9t² = 600(t-3) - 4.9(t-3)²
450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)
450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1
600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0
179.4t - 1844.1 = 0
t = (1844.1/179.4) = 10.28 s
Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.
y = 450t - 4.9t²
= (450×10.28) - (4.9×10.28×10.28)
= 4,108.2 m = 4.11 km
Hope this Helps
Ответ:
Well, do you have the length and width of the inclined plane? If not, there is no possible way to answer this. If you could add a photo of the question that would be great.
Explanation: