Question

if a sports car can go from rest to 27m/s in 9 sec. what is the magnitude of it'd average acceleration?

Answer 1

$a = \frac{v - u}{t}$
where a is acceleration, v is final velocity, u is initial velocity and t is time.
so your answer is
$a = \frac{27 - 0}{9} \\ a = \frac{27}{9} \\ a = 3$
average acceleration is 3m/s^2

Answer 2

hello dear user!
given,u=0m/s (as the object is initially at rest)
t=9s
v=27m/s
now,as we know that acceleration a=(v-u)/t
a=(27-0)/9=27/9=3m/s^2
thank u,hope it helps!