A car which is traveling at a velocity of 27 m/s undergoes an acceleration of 5.5 m/s^2 over a distance of 430 m. How fast is it going after that acceleration?

73.9m/s (1dp)

1) list everything that you are given using suvat, where s is distance, u is initial velocity(speed), v is final velocity(speed), a is acceleration and t is time

s = 430m

u = 27 m/s

v = ? (we need to work out)

a = 5.5m/s^2

t = (we are not given this value)

2) use an equation that doesn't involve the time

3) input the values that we have

so the answer is 73.9m/s to 1dp

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form