Answer:
[tex]-30m/s^{2}[/tex]
Explanation:
The first equation of motion given as,
v = u +at
Here, u is initial and v is final velocity and a is acceleration and t is time taken.
As per question car initial velocity, u = 90 m/s, final velocity, v =0 and time taken, t = 3.00 s.
Substituting these values in first equation of motion, we get
0 = 90 m/s + a×3.00 s
or, [tex]a = \frac{-90m/s}{3.00s}[/tex]
[tex]a=-30m/s^{2}[/tex]
Thus, the deacceleration of the car is [tex]-30m/s^{2}[/tex]
