# 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?

• u = 2ms-1
• v = 6ms-1
• t = 5s
• s = ?

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?

• u = ?
• v = 120ms-1
• t = 1min = 60s
• a = 5ms-2

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?

• u = 4ms-1
• a = 2ms-2
• t = 7s
• s = ?

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?

• u = 100ms -1
• v = 0 ( we want the plane to stop!)
• a = -5ms-2
• s = ?

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