Final answer:
The acceleration of the particle is shown to be perpendicular to the velocity at t=0 by examining their respective vector components. Since acceleration is solely in the negative x-direction without y or z components, any existing velocity in the y or z directions at t=0 would be perpendicular to the acceleration.
Step-by-step explanation:
To show that the acceleration of the particle at t=0 is perpendicular to the velocity, we should first understand that the velocity vector of the particle has components in the x, y, and z directions, with the x component being a linear function of time and the y and z components being constants. Taking the derivative of the velocity function with respect to time gives us the acceleration vector. Since the acceleration at a given time is -21 m/s² and constant in the negative x-direction, it does not have any components in the y or z directions. Thus, at t=0, if there is any velocity in the y or z directions (as the velocity is constant in these directions), it will be perpendicular to the acceleration, since the acceleration has no component in these directions.
Looking at the specific cases given, we see various instances where the velocity function and the acceleration are discussed. For example, at t=3s, the velocity is 15 m/s, and the acceleration is negative, indicating that the particle is slowing down. At t=5s, the velocity becomes negative, showing that the particle has changed direction. Moreover, at the initial moment, as the acceleration vector has no y or z components, and if the particle has initial velocity components in those directions, they would be perpendicular to the acceleration vector.
Ultimately, to conclusively show that the acceleration is perpendicular to the velocity at t=0, one would need the exact velocity vector components at that instant. Without those, we can only infer from the given information that if the particle has velocity components in the y or z direction at t=0, these components would be perpendicular to the acceleration vector.