Question

To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and position.

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 1.10 ft/s2

Part A - Finding the time interval for the motorcyclist to reach a given acceleration

If the motorcyclist starts from rest, determine the time needed to reach an acceleration of 5.25ft/s2

Part B - Speed of the Motorcyclist at a Given Acceleration

Determine the speed of the motorcyclist at the instant the acceleration is 6.75 ft/s2.

Part C - Magnitude of the Motorcyclist’s Acceleration at a Given Speed

Determine the magnitude of the acceleration at the instant the speed of the motorcyclist is 21.5 ft/s.

Part D - Maximum Constant Speed of the Motorcyclist

Determine the maximum constant speed of the motorcyclist when the maximum acceleration is 6.75 ft/s2.

Answer 1

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$

$a_t = 1.1ft/s^2$

PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $5.25ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$27.5625 = 1.21 + \frac{v^4}{122500}$

$v=42.3877ft/s$

Now calculate the angular velocity of the motorcycle

$v = r\omega$

$42.3877 = 350\omega$

$\omega = 0.1211rad/s$

Calculate the angular acceleration of the motorcycle

$a_t = r\alpha$

$1.1 = 350\alpha$

$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

$5.25ft/s^2$

$\omega = \alpha t$

$0.1211 = 3.1428*10^{-3}t$

$t = 38.53s$

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a_r = \frac{v^2}{R}$

$a_r = \frac{21.5^2}{350}$

$a_r =1.3207ft/s^2$

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a = \sqrt{a_t^2+a_r^2}$

$a = \sqrt{(1.1)^2+(1.3207)^2}$

$a = 1.7187ft/s^2$

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

Answer 2

(a) The time interval for the motorcyclist to reach a given acceleration is 38.5 s.

(b) The speed of the Motorcyclist at a given acceleration is 42.37 ft/s.

(c) The magnitude of the acceleration at the given instant speed is 1.72 ft/s².

(d) The maximum speed of the motorcycle at the given maximum acceleration is 42.28 ft/s.

The given parameters;

• radius of the curved path, R = 350 ft
• tangential acceleration, a = 1.1 ft/s²
• net acceleration, $a_t$ = 5.25 ft/s²

The centripetal acceleration of the motorcycle is calculated as;

$a_t = \sqrt{a^2 + a_c^2} \\\\a_t^2 = a^2 + a_c^2\\\\a_c^2 = a_t^2 - a^2 \\\\a_c = \sqrt{a_t^2 - a^2} \\\\a_c = \sqrt{(5.25)^2 - (1.1)^2} \\\\a_c =5.13 \ ft/s^2$

The linear velocity of the motorcycle is calculated as;

$a_c = \frac{v^2}{r} \\\\v^2 = a_c r\\\\v = \sqrt{a_c r} \\\\v = \sqrt{5.13 \times 350} \\\\v = 42.37 \ ft/s$

The time interval for the motorcyclist to reach a given acceleration;

$a = \frac{\Delta v}{\Delta t} \\\\\Delta t = \frac{\Delta v}{a} \\\\\Delta t = \frac{42.37}{1.1} \\\\\Delta t = 38.5 \ s$

The speed of the motorcycle at the acceleration is 6.75 ft/s2;

$a_c = \sqrt{a_t^2 - a^2} \\\\a_c = \sqrt{(6.75)^2 - (1.1)^2} \\\\a_c =6.66\ ft/s^2$

$a_c = \frac{v^2}{r} \\\\v^2 = a_c r\\\\v = \sqrt{a_c r} \\\\v = \sqrt{6.66 \times 350} \\\\v = 48.28 \ ft/s$

The magnitude of the acceleration at the instant the speed of the motorcyclist is 21.5 ft/s.

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(21.5)^2}{350} \\\\a_c = 1.32 \ ft/s^2\\\\a_t = \sqrt{a^2 + a_c^2} \\\\a_t = \sqrt{(1.1)^2 + (1.32)^2} \\\\a_t = 1.72 \ ft/s^2$

The maximum speed of the motorcycle when  the maximum acceleration is 6.75 ft/s².

v = 42.28 ft/s.

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