A remote-controlled car is moving in a vacant parking lot. The velocity of the c

Question

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v vector = [5.00 m/s - (0.0180 m/s^3) t^2]^i + [2.00 m/s + (0.550 m/s^2)t)^i. What is the magnitude of the acceleration of the car at t 7.59 s? Express your answer to three significant figures and include the appropriate units. What is the direction (in degrees counterclockwise from + x-axis) of the acceleration of the car at t = 7.59 s? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Given that -

v = [5.00 - (0.0180)t^2]i + [2.00 + (0.550)t]j

Part E -

Accele, a = dv/dt = (-2*0.0180t)i + (0.550)j

at t = 7.59 s

a = (-2*0.018*7.59)i + (0.550)j = (-0.273)i + (0.550)j

So, magnitude of acceleration = sqrt [(-0.273)^2 + (0.550)^2] = 0.6141 m/s^2

Answe in three significant digit = 0.614 m/s^2

Part F:

Angle with positive direction of x-axis = 180 - tan inverse(0.55/0.273) = 180 - 63.6 = 116.4 deg.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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