Answer to Question #236333 in Physics for eshiii

Question

 A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by 𝑣 = [5.00𝑚/𝑠 − (0.0180𝑚/𝑠 3 )𝑡 2 ]𝑖̂+ [2.00 𝑚⁄𝑠 + (0.550 𝑚 𝑠 2 ⁄ )𝑡]𝑗̂. (a) What are 𝑎𝑥(𝑡) and 𝑎𝑦(𝑡), the 𝑥- and 𝑦- components of the car’s velocity as functions of time? (b) What are the magnitude and direction of the car’s velocity at t = 8.00 s? (b) What are the magnitude and direction of the car’s acceleration at t = 8.00 s? 

Answer 1

a) The x and y componets of the car's velocity are respectively:

$$v_x(t) = 5.00m/s − (0.0180m/s^3 )t^2\\ v_y(t) = 2.00m/s − (0.550m/s^2 )t$$

The components of the acceleration are the corresponding derivatives:

$$a_x(t) = v'_x(t) = − 2\cdot (0.0180m/s^3 )t = -(0.036m/s^3 )t\\ a_y(t) = v'_y(t) = − 0.550m/s^2$$

b) The velocity vector at time $t =8s$ is given as follows:

$$\mathbf{v}(8s) =[ 5.00m/s − (0.0180m/s^3 )\cdot (8.00s)^2 ]\mathbf{i} + [2.00m/s − (0.550m/s^2 )\cdot 8s]\mathbf{j}=\\ =[3.848m/s]\mathbf{i} - [2.400m/s]\mathbf{j}$$

The magnitude is then:

$$v = \sqrt{3.848^2+2.400^2} \approx 4.54m/s$$

The direction is:

$$\theta_v =\arctan\left(\dfrac{-2.4}{3.848} \right) \approx 328\degree$$

c) Similarly for acceleration:

$$a =\sqrt{a_x^2(8s) + a_y^2(8s)} = \sqrt{(-0.288)^2 + (-0.550)^2} \approx 0.621m/s^2$$ $$\theta_a = \arctan\left(\dfrac{-0.550}{-0.288} \right) \approx 242\degree$$

Answer. a) See the solution. b) Magnitude: 4.54 m/s, direction: 328 degrees with positive x-axis. c) Magnitude: 0.621 m/s, direction: 242 degrees with positive x-axis.