The position of a squirrel running in a park is given by r rightarrow= [(0. 280m

Question

The position of a squirrel running in a park is given by r rightarrow= [(0. 280m/s)t + (0. 0360m/s2)/2]i^+ (0. 0190 m/s3)t3j^ What is upsilon x(t), the x-component of the velocity of the squirrel as function of time? upsilon x (t) = (0. 280 m/s)t + (0. 0720 m/s2)t2 upsilon x (t) = 0. 280 m/s upsilon x (t) = 0. 280 m/a + (0. 0720 m/s3)t upsilon x (t) = (0. 0720 m/s2)t What is upsilon v (t)- the y-component of the velocity of the squirrel as function of time? upsilon y (t) = (0. 0570 m/s3)t + (0,0720 m/s2)t2 upsilon y (t) = (0. 0570 m/s2)t3 upsilon y (t) = (0. 0570 m/s3)t3 upsilon y (t) = (0. 0570 m/s3)t At 4. 23s . how far is the squirrel from its initial position? Express your answer to three significant figures and include the appropriate units. At 4. 23s , what is the magnitude of the squirrel's velocity? Express your answer to three significant figures and include the appropriate units. At 4. 23s , what is the direction (in degrees counter clockwise from +x-axis) of the squirrel's velocity? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A) vx = drx/ dt = 0.28 + 0.072t m/s ans is c B) vy = dry/dt = 0.057 t^2 ans is c

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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