## Question

A motorcyclist heading east through a small Iowa city accelerates after he passes the signpost marking the city limits. His acceleration is a constant 4.0 ms^{-2}. At time t = 0 he is 5.0 m east of the signpost, moving east at 15 m/s.

(a) find his position and velocity at time t = 2.0 s.

(b) Where is the motorcyclist when his velocity is 25 m/s?

### Answer

**a) The answer to his position and velocity at time 2.0 seconds is 43 m and 23 m/s.**

**b) The location of the motorcyclist when his velocity is 25 m/s is 55 m.**

### Explanation

#### a)

**Data: Information from the question**

from kinematic equations

Distance is s_{1} = ut + (1/2)at^{2}

Additionally, we can also define distance as s_{2} = (x-x_{0})/t

When we assume s_{1} = s_{2}

**We can see that **

**ut + (1/2)at ^{2} = (x-x_{0})/t**

Therefore, we can now make x subject of the formula

x = x_{0} + ut + (1/2)at^{2}

and x_{0} = 5.0 m, u = 15 m/s, t = 2.0 s, a = 4.0 m/s^{2}

x = 5 + 15 x 2 + (1/2) x 4 x 2^{2} = 43 m

**Therefore, the position of the motorcyclist at t = 2.0 s is 43 meters (m).**

To find the velocity, we will use the equation v = u + at

Thus, v = 15 + 4 x 2 = 15 + 8 = 23 m/s

**Therefore, the velocity of the motorcyclist at time 2.0 s is 23 meters per second (23 m/s). **

#### b

We will use the equation v^{2} = u^{2} + 2a(x-x_{0})

And we will make x subject of the formula to obtain

x = x_{0} + (v^{2} – u^{2})/2a

By substituting the above equation with our data (remember that v = 25), we will now have

x = 5.0 + (25^{2} – 15^{2})/(2 x 4) = 55m

**Therefore, the position of the motorcyclist at v = 25 m/s is 55 meters (m).**

*You may also like to read:*