Which equation to use?

Which equation to use?Which equation to use?B R Sitaram,

Zeal Educational Services Pvt. Ltd.

Which Equation to use?Which Equation to use?

When students learn the equations of motion with constant acceleration, they often get confused: which equation should we use in a given situation?

This presentation shows a simple way of deciding!


First list the variables that we use:1.t: time2.s: distance3.u: initial speed4.v: final speed5.a: acceleration

Two more variables that can be removed easily:s0: initial position and t0: initial time.


Write the equations:1. s = u t + 1/2at2

2. v = u + a t3. v2-u2=2 a sMake a table, showing which are the variables used in each equation:


t s u v a

v = u + at Y N Y Y Y

s = ut+1/2at2 Y Y Y N Y

v2 – u2 = 2as N Y Y Y Y

Note: Each equation involves four of the five variables!Two equations appear to be missing: one not involving u and one not involving a.They are:s = vt – 1/2at2

s = t(u+v)/2


t s u v a

v = u + a t Y N Y Y Y

s = u t + ½ a t2 Y Y Y N Y

v2 – u2 = 2 a s N Y Y Y Y

s = v t - ½ a t2 Y Y N Y Y

s = t (u + v)/2 Y Y Y Y N


In all problems:4 variables will be referred to3 will be given, 1 to be determined

Use the table and determine which equation to use

Solve the equation!


Example:I throw a ball up with speed u against

gravity. How much time does it take to reach the top of its motion?

Variables: u, a (gravity), v (= 0 at top), t (to be determined)

Hence use v = u + a t


Example:I throw a ball up against gravity. It

takes 2 sec to reach the top of its flight. To what height does it go?

Variables: v (=0 at top), t, a (gravity), s (to be determined)

Hence, use s = v t – ½ a t2


Example:The Lamborghini Murcielago can

accelerate from 0 to 27.8 m/s (100 km/hr or 62.2 mi/hr) in a time of 3.40 seconds. Determine the acceleration of this car in both m/s/s and mi/hr/s.

Variables: u (=0), v (=27.8 m/s), t (3.40 s), a (to be determined)

Hence, use v = u + a t


Example:

A Formula One car is a single-seat racing car with an open cockpit and substantial wings located in the front and rear. At high speeds, the aerodynamics of the car help to create a strong downward force which allows the car to brake from 27.8 m/s (100 km/hr or 62.2 mi/hr) to 0 in as small of a distance as 17 meters. Determine the deceleration rate (i.e., acceleration) achieved by such a car.

Variables: u (=27.8 m/s), v (=0), s (17m), a (to be determined)

Hence, use v2 – u2 = 2as


Example:

A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

Variables: u (=0), v (=88.3 m/s), s (=1365 m), a (to be determined in 1st part), t (to be determined in 2nd part)

1st part: (u, v, s, a), use v2-u2=2as2nd part: (u, v, s, t), use s = t (u + v)/2(You can in fact use any equation for part 2, as 4

out of the 5 variables are now known!)

Which equation to use?

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