How long would it take a hawk to stop if he was flying at 9m/s and decelerated at 3m/s2?

3 seconds

Explanation:

From the question above,

we can apply the first equation of motion which is,

a = (v-u)/t Equation 1

Where a = acceleration. v = final velocity, u = initial velocity, t = time.

make t the subject of the equation

t = (v-u)/a Equation 2

Given: a = -3 m/s²(decelerating), u = 9 m/s, v = 0 m/s (to stop)

substitute these values into equation 2

t = (0-9)/-3

t = -9/-3

t = 3 seconds

a=d/t

a="triangle"v/t

explanation:

a = acceleration