An object moves such that its x-coordinate is given by x(t) = 6.0 m - (1.5 m/s2) t3 and its y-coordinate is given by y(t) = (8.0 m/s) t - (1.0 m/s2) t2. What is the acceleration of the object at t = 3.5 seconds? Express your answers in unit vector notation.
To find the acceleration of the object at t = 3.5 seconds, we need to differentiate the velocity equations with respect to time twice, as acceleration is the second derivative of position with respect to time.
Given:
x(t) = 6.0 m - (1.5 m/s^2) t^3
y(t) = (8.0 m/s) t - (1.0 m/s^2) t^2
First, let's find the velocity components by differentiating the position equations with respect to time:
vx(t) = dx(t)/dt = d/dt (6.0 m - (1.5 m/s^2) t^3)
= 0 - (1.5 m/s^2)(3t^2)
= -4.5 m/s^2 t^2
vy(t) = dy(t)/dt = d/dt ((8.0 m/s) t - (1.0 m/s^2) t^2)
= 8.0 m/s - (2.0 m/s^2) t
Next, let's find the acceleration components by differentiating the velocity equations with respect to time:
ax(t) = dvx(t)/dt = d/dt (-4.5 m/s^2 t^2)
= -9.0 m/s^2 t
ay(t) = dvy(t)/dt = d/dt (8.0 m/s - (2.0 m/s^2) t)
= -2.0 m/s^2
Now, let's substitute t = 3.5 seconds into these acceleration components to find the acceleration at t = 3.5 seconds:
ax(3.5) = -9.0 m/s^2 (3.5) = -31.5 m/s^2
ay(3.5) = -2.0 m/s^2
Therefore, the acceleration of the object at t = 3.5 seconds is (-31.5 m/s^2)î - (2.0 m/s^2)ĵ, where î and ĵ are unit vectors in the x and y directions respectively.