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. 0 Since this velocity has only j component, hence it will make an angle owith the x-axis. Again, we have v = 5t cm/s i + 5.00 cm/s j When t = 1.0 s v(1.0 s) = 5.00 cm/s i + 5.00 cm/s j 2 2 The magnitude of this velocity s = 5 + 5cm/s = 52 cm/s = 7.07 cm/s The angle made by this velocity vector with the x-component is tan 1= 5/5 , if 1 is the angle. tan = 1 1 0 1 45 We have v = 5t cm/s i + 5.00 cm/s j When t = 2.0 s v (2.0s) = 10.0 cm/s i + 5.00 cm/s j The magnitude of this velocity vector is 10 + 5 cm/s = 55 cm/s = 11.18 cm/s If 2 be the angle made by this velocity vector with the x-axis, tan = 5/10 2 = 26.56 0 2 0 So, the angle made by the above velocity vector with the x-axis is 26.56 . (c) The sketch of trajectory can be done as shown below.