CALC? A web page designer creates an animation in which a dot on a computer screen has position (a) Find the magnitude and direction of the dot’s average velocity between t = 0 and t = 2.0 s.(b) Find the magnitude and direction of the instantaneous velocity at t = 0, t = 1.0 s, and t = 2.0 s. (c) Sketch the dot’s trajectory from t = 0 to t = 2.0 s, and show the velocities calculated in part (b).

Solution 3E Given that Now, position vector at time t= 0s is r (0) = [4.0 cm + 0] i + 0 r (0) = 4.0 cm i Position vector at time t = 2.0 s is r (2.0) = [4.0 cm + 10.0 cm] i + 10 cm j r(2.0) = 14 cm i + 10 cm j r(2.0)r(0) Therefore, average velocity is =2.00 cm/s 14 cm i +10 cm j 4.0 cm j Average velocity = 2 s Average velocity = 5 cm/s i + 5 cm/s j (a) The magnitude of this velocity s = 5 + 5 cm/s = 52 cm/s Let be the angle made by the velocity vector with the x-axis. tan = 5/5 tan = 1 0 = 45 0 Therefore, the velocity vector makes an angle of 45 with the horizontal axis. (b) Differentiating the expression for r with respect to t, v = 5t cm/s i + 5.00 cm/s j When time t = 0, v (0) = 5.00 cm/s j Therefore, the magnitude of this velocity is 5.00 cm/s....