A car accelerates from rest at 2.3m/s^2 for 18 seconds, then decelerates to 15 m/s over 301 meters.

A) How fast is the car moving at 18seconds? (Is it 2.3 m/s^2)

B) How far does the car travel over the entire trip? (The m/s and m/s^2 is tripping me up)

C) How long does it take for the car to travel the entire trip?

D) What is the car's acceleration in the second part of the trip?

E) What is the car's average velocity for this trip?

Thank you for anyone who can help.

A.V=Vo + a*t = 0 + 2.3m/s^2*18s=41.4 m/s

B. d = d1 + d2
d = 0.5a*t^2 + 301 = 0.5*2.3*18^2 + 301
= 673.6 m.

C. V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = (15^2-41.4^2)/602 = -2.47 m/s^2 =
deceleration.

t2 = (V-Vo)/a = (15-41.4)/-2.47 = 10.7 s

t1+t2 = 18 + 10.7 = 28.7 s. for the entire trip.

D. a = -2.47 m/s^2.

E. V = d/(t1+t2) = 673.6/28.7=23.47 m/s.

To solve these questions, we can use the equations of motion. Let's go step by step:

A) To find the car's speed at 18 seconds, we can use the equation:
final speed = initial speed + (acceleration * time)

At rest, the initial speed is 0 m/s, the acceleration is 2.3 m/s^2, and the time is 18 seconds. Plugging these values into the equation, we get:
final speed = 0 m/s + (2.3 m/s^2 * 18 s) = 41.4 m/s

So, the car is moving at 41.4 m/s at 18 seconds.

B) To determine the total distance traveled by the car, we can break it into two parts: the acceleration phase and the deceleration phase.

In the acceleration phase, the car starts from rest and accelerates for 18 seconds. To find the distance covered during this phase, we can use the equation:
distance = (initial speed * time) + (0.5 * acceleration * time^2)

The initial speed is 0 m/s, the acceleration is 2.3 m/s^2, and the time is 18 seconds. Plugging these values into the equation, we get:
distance = (0 * 18) + (0.5 * 2.3 * (18^2)) = 369.9 meters

In the deceleration phase, the final speed is given as 15 m/s, and the distance covered is 301 meters.

So, the total distance traveled by the car over the entire trip is:
total distance = distance in acceleration phase + distance in deceleration phase
total distance = 369.9 meters + 301 meters = 670.9 meters

C) To find the time taken for the car to travel the entire trip, we can use the equation:
time = (final speed - initial speed) / acceleration

The initial speed is 0 m/s, the final speed is 15 m/s, and the acceleration is 2.3 m/s^2.
Plugging these values into the equation, we get:
time = (15 m/s - 0 m/s) / 2.3 m/s^2 = 6.52 seconds

So, it takes the car 6.52 seconds to travel the entire trip.

D) The car's acceleration in the second part of the trip (deceleration phase) is simply the negative of the acceleration in the first part. So, the acceleration in the second part is -2.3 m/s^2.

E) The average velocity for the whole trip can be calculated using the equation:
average velocity = total distance / total time

From the calculations above, we know the total distance is 670.9 meters, and the total time is 6.52 seconds. Plugging these values into the equation, we get:
average velocity = 670.9 meters / 6.52 seconds = 102.9 m/s

Therefore, the car's average velocity for this trip is 102.9 m/s.

I hope this explanation helps you understand how to solve these physics problems. Let me know if you have any further questions!