The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.?
The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.
At 4.69s , how far is the squirrel from its initial position?
At 4.69s , what is the magnitude of the squirrel's velocity?
At 4.69s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?
More Questions From This User See All
Answers & Comments
Verified answer
for the first one, just plug in 4.69 into the equation for t
.280*4.69 + .0360 * 4.69^2 + .0190* 4.69^3
For the second one plug 4.69 into the derivative
.280 + .720*4.69 + .57*.469^2
For the third one, I think you find the x y coordinates of the squirrels velocity and solve for the angle.
In what units is the first one in? What units are the second and third one in ?