The position of a squirrel running in a park is given by r =[(0.280m/s) t +(0.03

Question

The position of a squirrel running in a park is given by r =[(0.280m/s)t+(0.0360m/s2)t^2]i+ (0.0190m/s^3)t^3j.

Part C

At 5.25s , how far is the squirrel from its initial position?

Express your answer to three significant figures and include the appropriate units.

Part D

At 5.25s , what is the magnitude of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

Part E

At 5.25s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

The simplified equation is
r = (0.28t + 0.036t2)i + 0.019t3j

When t = 0, r = 0i + 0j. So the squirrel starts at the origin.

Part C:
When t = 5.25s, r = (0.28x5.25 + 0.036x5.252)i + 0.019x5.253j
r = 2.46i +2.75j

The distance = sqrt(2.462 + 2.752) = 3.69m

Part D:
Finding velocity by differentiation:
v = dr/dt
v = (0.28 + 0.072t)i + 0.057t2j

When t = 5.25s
v = (0.28 + 0.072x5.25)i + 0.057x5.252j
v = 0.658i + 1.571j

||v|| = sqrt (0.6582 + 1.5712) = 1.7m/s

Part E:
Direction Q = tan-1(1.571/0.658) = 67.2o

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.