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Hitman42 [59]
3 years ago
15

A body moves on a coordinate line such that it has a position sequals​f(t)equalstsquaredminus4tplus3on the interval 0less than o

r equalstless than or equals7​,with s in meters and t in seconds.a. Find the​ body's displacement and average velocity for the given time interval.b. Find the​ body's speed and acceleration at the endpoints of the interval.c.​ When, if​ ever, during the interval does the body change​ direction?
Physics
1 answer:
Zanzabum3 years ago
3 0

Answer:

A)  Δf = - 49 m, B)  v (7)  = -56 m / s, a = - 8 m / s², C)  t = 0.866 s

Explanation:

A) In this exercise ask to find the displacement and the average velocity, give the function of the movement

           f (t) = - 4t² +3

and the range of motion 0≤ t ≤ 7

the displacement is

for t = 0

           f (0) = 3

for t = 7 s

           f (7) = - 4 7² +3

           f (7) = -46 m

the total displacement is

           Δf = f (7) - f (0)

           Δf = -46 - 3

            Δf = - 49 m

the average speed is defined as the displacement between the time interval

           v = Df / Dt

           v = -49 / 7

           v = - 7 m / s

B) the speed and acceleration of the end points of the motion

         

the speed of defined by

          v = \frac{dx}{dt}

in this case

         v = \frac{df}{dt}

         v = -8t

         

let's calculate

          v (7) = -8 7

          v (7)  = -56 m / s

acceleration is defined by

          a = \frac{dv}{dt}

          a = - 8 m / s²

acceleration is constant throughout the movement

C) the point where the direction changes.

This point is a point where the position goes from positive to negative, the point f = 0

        0 = -4t² +3

        t = √¾

        t = 0.866 s

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MrMuchimi

Answer:

Explanation:

a) The formulas for the components Vx and Vy of the velocity and components x and y of the displacement are given by x = V0 cos(θ)       Vy = V0 sin(θ) - g t

x = V0 cos(θ) t       y = V0 sin(θ) t - (1/2) g t2

In the problem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2.

The height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero. That is when the projectile changes from moving upward to moving downward.(see figure above) and also the animation of the projectile.

Vy = V0 sin(θ) - g t = 0

solve for t

t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds

Find the maximum height by substituting t by 0.86 seconds in the formula for y

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b) The time of flight is the interval of time between when projectile is launched: t1 and when the projectile touches the ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence

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Solve for t

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two solutions

t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g

Time of flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.

c) In part c) above we found the time of flight t2 = 2 V0 sin(θ) / g. The horizontal range is the horizontal distance given by x at t = t2.

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d) The object hits the ground at t = t2 = 2 V0 sin(θ) / g (found in part b above)

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The magnitude V of the velocity is given by

V = √[ Vx2 + Vy2 ] = √[ (20 cos(25°))2 + (- V0 sin(25°))2 ] = V0 = 20 m/s

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