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?

A.) vy(t)=(0.0570m/s3)t+(0.0720m/s2)t2

B.) vy(t)=(0.0570m/s3)t2

C.) vy(t)=(0.0570m/s2)t2

D.) vy(t)=(0.0570m/s3)t

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?

Guest Aug 5, 2017

#1**+2 **

1. y(t) = 0.0190t^{3}

vy(t) = 3*0.0190t^{2} → 0.0570t^{2} If t is in seconds then the 0.0570 must have units m/s^{3} 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(X^{2} + Y^{2})

3. As for 2, but using vx(t) and vy(t) instead.

4. Calculate the angle from tan^{-1}(vy/vx)

Alan Aug 5, 2017