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Derivation of Distance to Get Velocity and Acceleration

Derivatives reveal a new connection between distance s(t), velocity v(t) and acceleration a(t). By differentiating, you can find an expression for velocity v(t) and acceleration a(t) using the provided distance s(t). This is how they are connected:

Rule

Distance, Velocity and Acceleration

If you have the distance (position) s(t), then it follows that

v(t) = s′(t) a(t) = v′(t) = s″ ⁡(t)

Example 1

A plane flies from London to New York. The function describes how far the aircraft has traveled

s(t) = 6000(3(t 8)2 −2(t 8)3)for 0 < t < 8

s(t) = 6000(3(t 8)2 − 2(t 8)3)for 0 < t < 8

where t is the number of hours after departure, and the unit of s(t) is kilometers.

1.
How far has the plane flown after four hours?
2.
What is the velocity of the plane after four hours?
3.
Find the acceleration of the plane after four hours.

1.
You insert t = 4 into s(t):
s(4) = 6000(3(4 8)2 − 2(4 8)3) = 3000

s(4) = 6000(3(4 8)2 − 2(4 8)3) = 3000

After four hours, the plane has flown 3000 km.

2.
To find the velocity of the plane after four hours, you have to find the velocity v(t) of the plane. You find that by differentiating s(t):
v(t) = s′(t) = −1125 16 t2 + 9000 16 t v(4) = s′(4) = −1125 16 ⋅ 42 + 9000 16 ⋅ 4 = 1125

v(t) = s′(t) = −1125 16 t2 + 9000 16 t v(4) = s′(4) = −1125 16 ⋅ 42 + 9000 16 ⋅ 4 = 1125

After four hours, the velocity of the plane is about 1125 km/h.

3.
To find out what the acceleration is after four hours, you have to differentiate the velocity function v(t). That gives you a(t) = v′(t) = s″ ⁡(t) = −1125 8 t + 4500 8 a(4) = v′(4) = s″ ⁡(4) = −1125 8 ⋅ 4 + 4500 8 = 0

After four hours, the acceleration of the plane is 0 km/h2. This tells you that the plane has zero acceleration at four hours into the flight.

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