The position of a particle moving in a two-dimensional plane is given by x = (3.

Question

The position of a particle moving in a two-dimensional plane is given by
x = (3.0 m/s)t - (4.0m/s2)t2
and
y = (6.0m/s^2)t^2 - 48t,
where x and y are in meters and t is in seconds.

a)Determine the magnitude and direction of the instantaneous acceleration at t = 3.0s.
b) Will the particle reach a maximum height? If so, what will be the maximum height?

Explanation / Answer

a) acceleration in x direction = d^2x/dt^2 = -8 m/sec^2 acceleration in y direction = d^2y/dt^2 = 12 m/sec^2 so acceleration = -8i+12j magnitude = sqrt(8^2 + 12^2) = 14.42 m/sec^2 direction = tan inverse( 12 / -8) = tan inverse( -4/3) with x axis. b)particle reach max height when x direction velocity =0 => dx/dt =0 => 3-8t =0 =>t =3/8 sec. max height = (vel iny) ^2 /2g vel in y direction = dy/dt = 12t -48 = 12*3/8 -48 = - 43.25 max height = 43.25*43.25 / 2*10 = 94.61 m

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