# A Cheetah is Crouched 20m to the East

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/s2)t2.

a) Find the displacement of the cheetah between t1 = 1.0 s and t2 = 2.0 s.

b) Find the average velocity during the same time interval

c) Find the instantaneous velocity at time t1 = 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 vx at t = 0.1 s and t = 0.2 s.

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

x = 20 m + (5.0 m/s2)t2

Thus, to find the displacement of the cheetah between t1 = 1.0 s and t2 = 2.0 s, we will apply the following method

At t1 = 1.0 s.

We will substitute t with t1 = 1.0 s in the equation x1 = 20 m + (5.0 m/s2)t12

Hence,

x1 = 20 m + (5.0 m/s2)(1.0)2 = 25 m

Similarly, we will apply the same method to find x2

Which implies

At t2 = 2.0 s

x2 = 20 m + (5.0 m/s2)t22 = 20 m + (5.0 m/s2)(2.0)2

Thus,

x2 = 40 m

To find the displacement between the two points, we will apply the equation x = x2 – x1

Since x2 = 40 m, and x1 = 25 m

We will now have x = x2 – x1 = 40 – 25 = 15 m

To find the average velocity, we will apply the formula

Average velocity (Vaverage) = Change in displacement / change in time

and

Change in displacement = x = x2 – x1 = 15 m

Also, Change in time = t = t2 – t1 = 2.0 – 1.0 = 1.0 s

Therefore,

Average velocity (Vaverage) = Change in displacement / change in time = 15 m / 1.0 s

And our final answer will be

Vaverage = 15 m/s

To find the instantaneous velocity at time t1 = 1.0 s.

At t = 0.1 s

Then the time interval from t1 to t2 is 1.1s where t1 = 1.0 s. Hence t2 = 1.1 s

Therefore, we will now apply the equation from the question to calculate x2

x2 = 20 m + (5.0 m/s2)t22

The above expression will now become

x2 = 20 m + (5.0 m/s2)(1.1)2 = 26.05 m

Now to calculate the average velocity, we will use x2 as 26.05 m in the equation

Average velocity (Vaverage) = Change in displacement / change in time = (x2 – x1) / (t2 – t1)

Hence,

Vaverage = (26.05 – 25) / (1.1 – 1.0) = 10.5 m/s

Also

At t = 0.01 s where t1 = 1.0 s

t2 = t1 + t = 1.0 + 0.01 = 1.01 s

x2 = 20 m + (5.0 m/s2)t22 = 20 m + (5.0 m/s2)(1.01 s)2 = 25.1005 m

Hence, Vaverage = (x2 – x1) / (t2 – t1) = (25.1005 – 25) / (1.01 – 1.0) = 10.05 m/s

At t = 0.001 s where t1 = 1.0 s

We will also have t2 = t1 + t = 1.0 + 0.001 = 1.001 s

Displacement, x2 = 20 m + (5.0)(1.001) = 20 m + (5.0)(1.001)2 = 25.010005 m

Thus, Vaverage = (x2 – x1) / (t2 – t1) = (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.

We will now apply a differential equation to solve question c

Remember, if we have xt2

Then dx/dt = 2xt2-1 = 2xt

We can equally say that

If x = 20 m + (5.0 m/s2)t2

Then dx/dt = (10.0 m/s2)t

Thus, the Vaverage = dx/dt = (10.0 m/s2)t

Hence, at t = 1.0 s

Vaverage = dx/dt = change in displacement / change in time = (10.0 m/s2)t = 10 x 1 = 10.0 m/s

Also, at t = 2.0 s

Vaverage = dx/dt = change in displacement / change in time = (10.0 m/s2)t = 10 x 2 = 20.0 m/s

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## Source

Cambridge university

A Cheetah is Crouched 20m to the East