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marissa [1.9K]
3 years ago
6

The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v

it+12at2xf=xi+vit+12at2 where xfxf is the final position, xixi is the initial position, vivi is the initial velocity, aa is the acceleration, and tt is the time. Let's say a car starts with an initial speed of 15 m/sm/s, and moves between the 1000 mm and 5000 mm marks on a roadway in a time of 60 ss. What is its acceleration?
Physics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

a = -04978 m / s²

Explanation:

For this exercise we can use the kinematics equations, as the initial speed, distance and time indicate, we can use

        x = x₀ + v₀ t + 1 /2 a  t²

       ½ a t² = x-x₀ - v₀ t

       a = (x-x₀ - v₀ t) 2 / t²

let's reduce the magnitudes to the SI system

     x₀ = 1000 mm = 1 m

     x  = 5000mm = 5m

let's calculate

     a = (5 - 1 - 15 60) 2/60²

     a = -04978 m / s²

negative sign indicates that braking is related

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A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
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Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

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