Question

The acceleration of a particle is given by ax1t2 = -2.00 m>s 2 13.00 m>s 3 2t. (a) find the initial velocity v0x such that the particle will have the same x-coordinate at t = 4.00 s as it had at t = 0. (b) what will be the velocity at t = 4.00 s?

Answer 1

To find the initial velocity v₀x such that the particle will have the same x-coordinate at t = 4.00 s as it had at t = 0, integrate the acceleration function ax₁t² = -2.00 m/s² + 13.00 m/s³t² + 2t with respect to time to get the velocity function vx₁t = -2.00t + 4.3333t³ + t². Then, set x(t=4.00) = x(t=0) and integrate the velocity function to find v₀x = -42.667 m/s. To find the velocity at t = 4.00 s, substitute t = 4.00 s into the velocity function and solve to get v = 134.667 m/s.

Step-by-step explanation:

To find the initial velocity v₀x such that the particle will have the same x-coordinate at t = 4.00 s as it had at t = 0, we need to integrate the acceleration function ax₁t² = -2.00 m/s² + 13.00 m/s³t² + 2t with respect to time to get the velocity function vx₁t = -2.00t + 4.3333t³ + t². Setting x(t=4.00) = x(t=0) and integrating the velocity function with respect to time, we can find v₀x = -42.667 m/s.

To find the velocity at t = 4.00 s, we can substitute t = 4.00 s into the velocity function vx₁t = -2.00t + 4.3333t³ + t² to get v = -2.00(4.00) + 4.3333(4.00)³ + (4.00)² = 134.667 m/s.