A cheetah is crouched 20m to the east of an observer’s vehicle. At time t = 0 the cheetah charges an antelope and begins to run along a straight line. During the first 2.0 seconds of the attack, the cheetah’s coordinate x varies with the time. This is according to the equation x = 20 m + (5.0 m/s^{2})t^{2}.

a) Find the displacement of the cheetah between t_{1} = 1.0 s and t_{2} = 2.0 s.

b) Find the average velocity during the same time interval

c) Find the instantaneous velocity at time t_{1} = 1.0 s. By taking ∆t = 0.1 s, then ∆t = 0.o1 s, then ∆t = 0.001 s.

d) Derive a general expression for the instantaneous velocity as a function of time. Also from it find v_{x} at t = 0.1 s and t = 0.2 s.

### Answer to a

We need to first remember that we were given an equation:

x = 20 m + (5.0 m/s^{2})t^{2}

Thus, to find the displacement of the cheetah between t_{1} = 1.0 s and t_{2} = 2.0 s, we will apply the following method

**At t _{1} = 1.0 s. **

We will substitute t with t_{1} = 1.0 s in the equation x_{1} = 20 m + (5.0 m/s^{2})t_{1}^{2}

Hence,

x_{1} = 20 m + (5.0 m/s^{2})(1.0)^{2} = 25 m

Similarly, we will apply the same method to find x_{2}

Which implies

**At t _{2} = 2.0 s**

x_{2} = 20 m + (5.0 m/s^{2})t_{2}^{2} = 20 m + (5.0 m/s^{2})(2.0)^{2}

Thus,

x_{2} = 40 m

To find the displacement between the two points, we will apply the equation x = x_{2} – x_{1}

Since x_{2} = 40 m, and x_{1} = 25 m

We will now have x = x_{2} – x_{1} = 40 – 25 = 15 m

### Answer to b

To find the average velocity, we will apply the formula

Average velocity (V_{average}) = Change in displacement / change in time

and

Change in displacement = x = x_{2} – x_{1} = 15 m

Also, Change in time = t = t_{2} – t_{1} = 2.0 – 1.0 = 1.0 s

Therefore,

Average velocity (V_{average}) = Change in displacement / change in time = 15 m / 1.0 s

And our final answer will be

V_{average} = 15 m/s

### Answer to c

To find the instantaneous velocity at time t_{1} = 1.0 s.

At t = 0.1 s

Then the time interval from t_{1} to t_{2} is 1.1s where t_{1} = 1.0 s. Hence t_{2} = 1.1 s

Therefore, we will now apply the equation from the question to calculate x_{2}

x_{2} = 20 m + (5.0 m/s^{2})t_{2}^{2}

The above expression will now become

**x _{2 }= 20 m + (5.0 m/s^{2})(1.1)^{2} = 26.05 m**

Now to calculate the average velocity, we will use x_{2} as 26.05 m in the equation

Average velocity (V_{average}) = Change in displacement / change in time = (x_{2} – x_{1}) / (t_{2} – t_{1})

Hence,

**V _{average} = (26.05 – 25) / (1.1 – 1.0) = 10.5 m/s**

Also

**At t = 0.01 s where t _{1} = 1.0 s**

t_{2} = t_{1} + t = 1.0 + 0.01 = 1.01 s

x_{2} = 20 m + (5.0 m/s^{2})t_{2}^{2} = 20 m + (5.0 m/s^{2})(1.01 s)^{2} = 25.1005 m

Hence, V_{average} = (x_{2} – x_{1}) / (t_{2} – t_{1}) = (25.1005 – 25) / (1.01 – 1.0) = 10.05 m/s

**At t = 0.001 s where t _{1} = 1.0 s**

We will also have t_{2} = t_{1} + t = 1.0 + 0.001 = 1.001 s

Displacement, x_{2} = 20 m + (5.0)(1.001) = 20 m + (5.0)(1.001)^{2} = 25.010005 m

Thus, V_{average} = (x_{2} – x_{1}) / (t_{2} – t_{1}) = (25.010005 – 25) / (1.001 – 1.0) = 10.005 m/s

The more the size t reduces, the more the average velocity reduces to 10 m/s.

**Therefore, we can conclude that at t = 1.0 s, the average velocity is 10.0 m/s.**

### Answer to d

We will now apply a differential equation to solve question c

Remember, if we have xt^{2}

Then dx/dt = 2xt^{2-1} = 2xt

We can equally say that

If x = 20 m + (5.0 m/s^{2})t^{2}

Then dx/dt = (10.0 m/s^{2})t

Thus, the V_{average} = dx/dt = (10.0 m/s^{2})t

Hence, at t = 1.0 s

V_{average} = dx/dt = change in displacement / change in time = (10.0 m/s^{2})t = 10 x 1 = 10.0 m/s

Also, at t = 2.0 s

V_{average} = dx/dt = change in displacement / change in time = (10.0 m/s^{2})t = 10 x 2 = 20.0 m/s

*You may also like to read*

## Source

A Cheetah is Crouched 20m to the East