After a ball rolls off the edge of a horizontal table at a time t = 0, its veloc

Question

After a ball rolls off the edge of a horizontal table at a time t = 0, its velocity as a function of time is given by V^rightarrow = 1.0icirc - 9.8jcirc Where V^rightarrow is in meters per second and t is in seconds. The ball's displacement away from the edge of the table, during the time interval of 0.365 s for which the ball is in flight, is given by Deltar^rightarrow = integral_0^0.365 V^rightarrow dt. To perform the integral, you can use the calculus theorem integral (A + B f(x))dx = integral A dx + B integral f(x) dx. You can think of the units and unit vectors as constants, represented by A and B. perform the integration to calculate the displacement of the ball from the edge of the table at 0.365 s. delta r^rightarrow = __________ m

Explanation / Answer

Basically the vector has a horizontal component, i, which has a constant velocity of 1.0 m/s; and a vertical component, j, which has an accelerating velocity -9.8*t

The integral of the vector is displacement
d(t, i, j)=1.0*t i-.5*9.8*t^2 j

To find the displacement in 0.365 seconds, plug that into the vector equation.

Look at
d(t=0,i,j) versus d(t=0.365,i,j)

d(t=0,i,j)=(0i,0j)

d(t=0.365,i,j)=1.0*0.365i, -.5*9.8*0.365^2j

=0.365i, -0.652j

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