Question

Water rushes down a river with a velocity of 2.4 m/s. Four seconds after leaving the rapids the velocity of the water has decreased to just 0.6 m/s. What is the deceleration of the water?

Answer 1

`-0.45ms^-2`

Explanation:

SUVat

U = 2.4 - initial velocity
V = 0.6 - Final Velocity
t = 4 Time
we're after a Acceleration

One of the equations we can use is

`V = U+at`

sub in your numbers
`0.6 = 2.4 + a(4)`
bring the 2.4 over

`-1.8 = 4a`

• divide both sides by 4

`-1.8/4 = a`
Which equals `-0.45ms^-2`
Don't be afraid of the negative, it's decelerating remember ;)