1. 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^.
What is υy(t), the y-component of the velocity of the squirrel, as function of time?
2. At 4.52 s , how far is the squirrel from its initial position?
3. At 4.52 s , what is the magnitude of the squirrel's velocity?
4. At 4.52 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?
1. y(t) = 0.0190t3
vy(t) = 3*0.0190t2 → 0.0570t2 If t is in seconds then the 0.0570 must have units m/s3 as the result must be a velocity (m/s). Hence B.) is the result.
2. Put t = 4.52 s into x(t) from your previous question to find X, the distance travelled in the x-direction.
Put it into y(t) above to find Y, the distance travelled in the y-direction.
Then calculate R = sqrt(X2 + Y2)
3. As for 2, but using vx(t) and vy(t) instead.
4. Calculate the angle from tan-1(vy/vx)