4.05 suvat examples

Example 1
An object accelerates over a period or 5 seconds from 2ms -1 to 6ms -1
How far does it travel during this time?

Use equation (1)
s = (v + u)/2 x t
s = (6+2)/2 x 5 = 20m

Example 2
An object reaches a final velocity of 120ms -1 after accelerating at a rate of 5ms -2 for 1 minute
What was its initial velocity?

Use equation (2)
v = u + at     Therefore u = v -at
u = 120 - 5 x 60 = -180ms-1
Note this means the object was initially travelling in the opposite direction at 180ms -1

Example 3
An object that is initially travelling at 4ms-1accelerates at a rate of 2ms -2 over a period or 7 seconds
How far does it travel during this time?

Use equation (3)
s = ut + 1/2at2
s = 4 x 7 + 1/2 x 2 x 72 = 77m

Example 4
An plane lands on a runway with an initial velocity of 100ms -1 and the maximum acceleration produced by the brakes is -5ms -2 (note the acceleration due to the breaks is opposite to the direction of velocity!)
What is the minimum length of runway needed for the plane to come to rest?

Use equation (4)
v2 = u2 + 2as     Therefore s = (v 2 - u 2)/2a
s = 0 - 100 2/(2 x (-5)) = 1000m (1km)