Final answer:
The question involves calculating the velocity and acceleration of a particle at t = 1 second based on given position equations. By differentiating these equations with respect to time, the specific values at the instant can be determined.
Step-by-step explanation:
The student's question involves determining the velocity and acceleration of a particle at a certain time given its motion equations in the x and y directions. The particle's motion is defined by the equations x = t³ - 2t² and y = 2t, where x and y are in meters and t is in seconds. We have to find velocity and acceleration when t = 1 second.
To calculate the velocity, we need to determine the time derivatives of the position functions for each axis. The velocity components will be vx (t) = dx/dt and vy (t) = dy/dt. Similarly, acceleration components will be ax (t) = d2x/dt2 and ay (t) = d2y/dt2. After taking the derivatives, we substitute t = 1 to find the specific values for velocity and acceleration at that time.
To determine the velocity and acceleration when t = 1 second, we need to find the derivatives of the equations for x and y with respect to t.
The derivative of x = t³ - 2t² is obtained by applying the power rule: dx/dt = 3t² - 4t. Substituting t = 1 into this equation gives us the velocity at t = 1 second. Therefore, v = 3(1)² - 4(1) = -1 m/s.
The derivative of y = 2t is dy/dt = 2. The acceleration is the derivative of the velocity, so a = d²x/dt² = d(3t² - 4t)/dt = 6t - 4. Substituting t = 1 into this equation gives us the acceleration at t = 1 second. Therefore, a = 6(1) - 4 = 2 m/s².