0.5 Quadratic equations  (Page 2/5)

Because the dragster is initially motionless, its initial velocity ( v ₀ ) is 0. Also since the dragster is situated at the starting point at time t = 0, its initial position ( x ₀ ) is 0. We substitute these values into (2) to obtain

We can determine the distance that the dragster travels in 1 second by substituting the value 1 for t .

Thus we conclude that the dragster travels 25 meters in its first second of travel.

Example 2: How far will the dragster travel in 2 seconds?

We may make use of the same equation (2) as before. In this case, we substitute 2 for t to obtain the distance traveled by the dragster in 2 seconds.

Example 3: Most drag strips are one-quarter mile in length. Assuming that the acceleration remains uniform, how long will it take the dragster to complete the one-quarter mile strip?

Because the equation of motion for the dragster is written is MKS units, we must convert one-quarter mile to meters. We can do so by applying the unit conversion factor 1 mile = 1,609.344 m. Thus,

Once again, equation (2) can be used to model the motion of the dragster. We will use this equation to solve for the amount of time ( t ₁ ) that it will take the dragster to travel 402.3 meters. The steps are shown below.

So we conclude that it will take the dragster slightly 4.01 seconds to complete the quarter mile track.

Projectile motion

Just as we could model a dragster as an object under constant acceleration, we can also model a projectile using a similar approach. A projectile is an object that is hurled into the air. Rockets, missiles and artillery shells are examples of projectiles.

Projectiles experience uniform acceleration as they travel through the sky. The uniform acceleration is due to the gravitational pull exerted on the projectile by the Earth. This acceleration is denoted as g . It value in the MKS system is 9.8 m / s ² . In the British system of units, the value of g is equal to 32 ft/s ² .

Let us represent the height of a projectile as a function of time as y (t). Let us represent the initial velocity in the vertical direction of the projectile and the initial height of the projectile to be v ₀ and y ₀ , respectively. We can incorporate these constants into equation (1) to develop the following equation of motion for a projectile.

The presence of the minus sign in the equation of motion for a projectile is due to the fact that the gravitation force exerted on the projectile is directed toward the Earth which is opposite to the direction of increasing vertical position.

We can use this equation to model the height of a projectile as a function of time. The following exercises illustrate this principle. In the next three exercises you can ignore any affects due to the atmosphere or wind.

Example 4: Let us consider a bottle rocket that is launched at time ( t = 0) from a flat surface. The initial velocity of the bottle rocket is 124 ft/s. Determine the height of the bottle rocket 1.5 seconds after it is launched.