Uniformly Accelerated Motion

Uniformly accelerated motion is motion with a constant acceleration. It is motion where the rate of change of velocity with time is constant. Since the acceleration is a constant, average and instantaneous acceleration are the same. Thus, since a_av = ( v_f - v_i) ⁄ t (Assuming t_i = 0 we may write Δt = t ), a = (v_f - v_i) ⁄ t. Or

v_f = v_i + at

Since the velocity changes uniformly, the average velocity of a uniformly accelerated motion is equal to the average of its initial and final velocities.

v_av = ( v_i + v_f ) ⁄ 2

and since Δx = v_av t,

Δx = ( v_i + v_f )( t ⁄ 2)

Two other equations can be obtained by substituting for v_f and for t into this equation. Replacing v_f by ( v_i + at ), The following equation can be obtained.

Δx = v_i t + (a ⁄ 2)t²

And replacing t by ( v_f - v_i) ⁄ a, The following equation can be obtained.

v_f ² = v_i² + 2aΔx

The following is a list of the four equations of a uniformly accelerated motion:

v_f = v_i + at

Δx = v_it + ( a ⁄ 2)t²

v_f ² = v_i² + 2aΔx

Δx = (v_i + v_f )( t ⁄ 2)

Only two of these equations are independent. These equations involve five variables: t , a , Δx , v_i , and v_f. Since only two of these equations are independent, if any three of these variables are known, we can usually solve for the other two by using suitable equations.

Example: The speed of a car changed from 20 m ⁄ s to 40 m ⁄ s in 10 seconds.

1. Calculate its acceleration.

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   Solution: t = 10 s; v_i = 20 m ⁄ s; v_f = 40 m ⁄ s; a = ?

   v_f = v_i + at

   a = ( v_f - v_i ) ⁄ t = ( 40 - 20 ) ⁄ 10 m ⁄ s² = 2 m ⁄ s²

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2. Calculate the distance travelled.

   Solution:

   Δx = v_i t + (a ⁄ 2)t² = {(20)(10) + (2 ⁄ 2)(10²)} m = 300 m.

Example: A car initially moving with a speed of 50 m ⁄ s was stopped in a distance of 100 m.

1. Calculate its acceleration.

   Solution: v_i = 50 m ⁄ s; v_f = 0 ; Δx = 100 m; a = ?

   v_f ² = v_i² + 2aΔx

   a = ( v_f ² - v_i² ) ⁄ ( 2Δx ) = (0² - 50² ) ⁄ 2 ⁄ Δx ) m ⁄ s² = -12.5 m ⁄ s²

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2. Calculate the time taken.

   Solution:

   v_f = v_i + at

   t = (v_f - v_i) ⁄ a = (0 - 50)/(-12.5) s = 4 s.

Motion under Gravity

Motion under gravity is a uniformly accelerated motion. The absolute value of gravitational acceleration ( g ) is 9.8 m ⁄ s². Since the effect of gravitational acceleration is to decrease velocity (decrease a positive velocity when going up and increase a negative velocity when coming down), gravitational acceleration is negative.

g = -9.8 m ⁄ s²

Equations of Motion under Gravity

Since gravitational motion is a uniformly accelerated motion, its equations can be obtained from the equations of a uniformly accelerated motion by replacing a by g. Also, since gravitational motion is a vertical motion, Δx needs to be replaced by Δy. Δy is taken to be positive if the final position is above the initial position and negative if the final position is below the initial position. The following are equations of gravitational motion.

v_f = v_i + gt

Δy = v_it + ( g ⁄ 2)t²

v_f ² = v_i² + 2gΔy

Δy = ( v_i + v_f )(t ⁄ 2)

These equations involve four variables. If any two of these variables are known, we can usually solve for the other variables.

1. Calculate the time taken to reach the ground.

   Solution: v_i = 0 (because it is dropped from rest); Δy = -20 m (negative, because it is going down); t = ?

   Δy = v_i t + ( g ⁄ 2 )t² = ( g ⁄ 2)t² (since v_i = 0 m ⁄ s)

   t = √( 2Δy ⁄ g ) = √{2(-20)/(-9.8)} s ≈ 2 s

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2. Calculate its speed by the time it hits the ground.

   Solution: v_f = ?

   v_f = v_i + at = {0 + (-9.8)(2)} m ⁄ s ≈ = -20 m ⁄ s

Example: A ball is thrown upward with a speed of 20 m ⁄ s.

1. Calculate the time taken to reach the maximum height.

   Solution: v_i = 20 m ⁄ s; v_f = 0 m ⁄ s (At maximum height speed is zero).

   v_f = v_i + at

   t = ( v_f - v_i)/g = (0 - 20)/(-9.8) ≈ 2 s

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2. To what height would it rise?

   Solution: Δy = ?

   Δy = (v_i + v_f )(t ⁄ 2) = (20 + 0)(2 ⁄ 2) m ≈ 20 m

Example: A ball is thrown upwards from a 10 m tall building with a speed of 10 m ⁄ s.

1. Calculate the speed with which it will hit the ground .

   Solution: v_i = 10 m ⁄ s; Δy = -10 m (negative because the final position is below the initial position); v_f = ?

   v_f² = v_i² + 2gΔy

   v_f = √{ v_i² + 2gΔy} = - √{ 10² + 2(-9.8)(-10)} m ⁄ s ≈ -17.3 m ⁄ s

   The negative square root is taken because it is going down.

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2. Calculate the time taken to hit the ground.

   Solution: t = ?

   v_f = v_i + gt

   t = ( v_f - v_i ) ⁄ g = (-17.3 - 10)/(-9.8) s ≈ 2.73 s