Question

You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, the car is at r⃗ 0=(3.0i^+2.0j^)m.

What is the y component of the car's position vector at 8.0 s ?
What is the x component of the car's acceleration vector at 8.0 s ?
What is the y component of the car's acceleration vector at 8.0 s ?

Answer 1

To find the y component of the car's position vector at 8.0 s, we need to use the y component of the velocity vector and the y component of the initial position vector. The y component of the velocity vector is 2t^2 j^. So at t = 8.0 s, the y component of the position vector is:
r⃗y=(r⃗0y)+(v⃗yt)=2*(8.0s)^2 j^ = 256 j^
To find the x component of the car's acceleration vector at 8.0 s, we need to find the derivative of the x component of the velocity vector with respect to time. The x component of the velocity vector is -3ti^. So the derivative of the x component of the velocity vector is:
a⃗x=-3i^
To find the y component of the car's acceleration vector at 8.0 s, we need to find the derivative of the y component of the velocity vector with respect to time. The y component of the velocity vector is 2t^2 j^. So the derivative of the y component of the velocity vector is:
a⃗y=4tj^
at t=8s, a⃗y=4*8j^=32j^