A sports car moving at constant speed travels 150 meters in 5.4 seconds , then brakes and comes to a stop in 3.9 seconds

What is the magnitude of its acceleration in ?

What is the magnitude of its acceleration in g's (g= 9.80 m/s^2 )?

initial velocty=150/5.4 m/s

a= changevelocity/time=Vi/3.9

a=(Vf-Vi)/t

Vf=0 (it stopped moving)
Vi=(150/5.4)m/s=28m/s
total t = 9.3s
so
a=(0-28)/9.3 = 3 m/s

To find the magnitude of acceleration, we need to use the formula for acceleration, which is:

acceleration = (change in velocity) / (time taken)

First, let's find the change in velocity during the first part of the motion when the car travels 150 meters in 5.4 seconds.

change in velocity = (final velocity) - (initial velocity)

Since the car is moving at constant speed, the initial velocity is the same as the final velocity. Therefore, the change in velocity is 0 m/s.

Now, let's calculate the acceleration during the first part of the motion:

acceleration = 0 m/s / 5.4 s
= 0 m/s^2

During the second part of the motion, the car comes to a stop in 3.9 seconds. In this case, the change in velocity is the final velocity of 0 m/s minus its initial velocity.

change in velocity = 0 m/s - (initial velocity)

Since the car comes to a stop, the initial velocity is not given, but we know that it is a positive value (since the car is moving in the positive direction). Therefore, we can say:

acceleration = -initial velocity / 3.9 s
= 0 m/s^2 - initial velocity / 3.9 s

Now, let's calculate the acceleration in g's. To do this, we need to divide the acceleration in m/s^2 by the value of g, which is 9.80 m/s^2.

acceleration in g's = acceleration / g
= (-initial velocity / 3.9 s) / 9.80 m/s^2

Please note that we were not given the value of the initial velocity, so we cannot determine the exact magnitude of acceleration or acceleration in g's without that information.