Question

A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1300 N.What is the magnitude of the car's acceleration at t=13s?What is the direction of the car's acceleration at t=13s ? Give the direction as an angle from the r-axis.If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?

Answer 1

Answer:

$The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad$

Explanation: