A motorcyclist travels around a curved path that has a radius of 350 ft. While traveling around the curved path, the motorcyclist increases speed by 2 ft/s2 . A. If the motorcyclist starts from rest, determine the time needed to reach a total acceleration of 10 ft/s2 . B. Determine the speed of the motorcyclist at the instant with the total acceleration of 10 ft/s2 . C. Determine the magnitude of the total acceleration at the instant when the speed of the motorcyclist is 20 ft/s

A) t = 29.3 s

B) V = 58.56 ft/s

C) a_net = 2.3 ft/s²

Explanation:

A) The formula for radial acceleration is given as;

a_c = V²/R

We are given;

Radius;R = 350 ft

So, a_c = V²/350

Where V is velocity

Tangential acceleration;a_t = 2 ft/s²

Formula for net acceleration is;

a_net = √((a_c)² + (a_t)²)

We are given a_net = 10 ft/s²

Thus;

10 = √(V²/350)² + (2²)

10² = V⁴/350² + 4

100 - 4 = V⁴/350²

96 × 350² = V⁴

V = 58.56 ft/s

Now, formula for angular velocity is;

ω = V/r

ω = 58.56/350

ω = 0.1673 rad/s

Angular acceleration is given by;

α = a_t/r

α = 2/350

α = 0.00571 rad/s²

Time needed will be gotten from the formula;

t = ω/α

t = 0.1673/0.00571

t = 29.3 s

B) we are told total acceleration is 10 ft/s², thus it's the same as velocity gotten earlier which is 58.56 ft/s

C) we are told that the speed is now 20 ft/s

Thus;

a_c = 20²/350

a_c = 1.1429 ft/s²

Since a_net = √((a_c)² + (a_t)²)

We are given a_t = 2 ft/s²

Thua;

a_net = √(1.1429² + 2²)

a_net = √5.30622041

a_net = 2.3 ft/s²

friction

explanation:

they us to move forward and backwards

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank