## Question

The motorcycle is traveling at a constant speed of 76km/h. Determine the magnitude of its acceleration when it is at point A.

## Answer

**The magnitude of its acceleration when it is at point A is 1.223 m/s ^{2}. **

### Explanation

**Data:** The information from the question

Constant speed, v = 76 km/h = (76 x 1000) / (60 x 60) = 76000 / 3600 = 21.1 m/s

The tangential acceleration, a_{t} = dv/dt = 0

x = 25 m

We also have y^{2} = 2x. Thus, we can take the square root of both sides to end up with y = √2x

But since x = 25, we can then say that y = √2x = √2 x 25 = √50. Therefore, y = √50.

**Unknown:** The information we need to find

The normal acceleration, a_{n} = ?

**Formula: **The equation that will help us to solve the problem

First step: We will apply the first and second derivatives of 2x

Second Step: We will use the product rule

Third step: We will now employ the equation for the radius of the curvature **ρ = [1 + (dy/dx) ^{2}]^{3/2} / (d^{2}y/dx^{2})**

Fourth Step: We will finally apply the formula **a _{n} = v^{2} / ρ**

#### Solution

**First step:**

y^{2} = 2x

After differentiating the above equation, we will now have

2y (dy/dx) = 2

We can now divide both sides by 2y to obtain

(dy/dx) = 1/y

**Second step:**

We will now use the product rule

2y (d^{2}y/dx^{2}) + 2 (dy/dx)^{2} = 0

After collecting like terms and dividing both sides by 2y, we will get:

(d^{2}y/dx^{2}) = – (1/y)(dy/dx)^{2}

We can now substitute our data into the above equation

(d^{2}y/dx^{2}) = – (1/√50)(dy/dx)^{2}

But dy/dx = 1/√50

Therefore,

(d^{2}y/dx^{2}) = – (1/√50)(1/√50)^{2}

We will now have

(d^{2}y/dx^{2}) = (1/(50√50))

(d^{2}y/dx^{2}) = (1/353.6) = 0.00283 = 2.83 x 10^{-3}

Remember that dy/dx 1/√50 = 0.141

**Third step:**

We will now use the formula **ρ = [1 + (dy/dx) ^{2}]^{3/2} / (d^{2}y/dx^{2})** to find the radius of the curvature

**(ρ).**

**ρ = [1 + (dy/dx) ^{2}]^{3/2} / (d^{2}y/dx^{2}) = [1 + (0.141)^{2}]^{3/2} / (2.83 x 10^{-3})** =

**363.95**

**m**

**Fourth step:**

The final step is to use the formula **a _{n} = v / ρ** to find the magnitude of the acceleration. Therefore

** a_{n} = v^{2} / ρ** =

**(21.1 m/s)**

^{2}/

**(363.95 m)**

**= 445.21 m**

^{2}/s^{2}/ 363.95 m**= 1.223 m/s**

^{2}**Therefore, the magnitude of the acceleration is 1.223 meters per second square (m/s ^{2})**

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