Applied Mechanics: Dynamics: Spring 2018 Tyler Heath Chapter 11A instructions I

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Applied Mechanics: Dynamics: Spring 2018 Tyler Heath Chapter 11A instructions I hel Question 1 (of 9) Save &Exit; Submit 1 1.00 points value: The motion of a particle is defined by the relation x-6-23-1223 3, where x and t are expressed in m/s2 meters and seconds, respectively. Determine the time, position, and velocity when acceleration a 2 The time, position, and velocity when acceleration a 2 m/s2 are S, m, and m/s, respectively. Hints References eBook & Resources Hint#1 Check my work D FS FS F6 8

Explanation / Answer

Given,

x = 6t^4 - 2t^3 - 12t^2 + 3t + 3

we have, a = 2 m/s^2

We need to know the time at which a = 2 m/s^2 to find position and velocity

We know that,

a = dv/dt = d2x/dt^2

v = dx/dt = d(6t^4 - 2t^3 - 12t^2 + 3t + 3)

v = 24 t^3 - 6t^2 - 24t + 3

a = dv/dt = d(24 t^3 - 6t^2 - 24t + 3)/dt

a = 72 t^2 - 12 t - 24

72 t^2 - 12t - 24 = 2

72t^2 - 12t - 26 = 0

the above quadratic eqn gives us:

t = 0.69, -0.52

taking the positive value

x = 6 x 0.69^4 - 2 x 0.69^3 - 12 x 0.69^2 + 3 x 0.69 + 3 = 0.0598 m

v = 24 x 0.69^3 - 6 x 0.69^2 - 24 x 0.69 + 3 = -8.53 m/s

Hence, t = 0.69 s ; x = 0.059 m ; v = -8.53 m/s

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