Motion in 1 dimension easy question?

In summary, a car starts from rest and travels for 5.0 seconds with a uniform acceleration of +1.5m/s^2. The driver then applies the brakes, causing uniform acceleration of -2.0m/s^2. The car is going at 7.5 m/s when it begins to brake and after 3.0 seconds, it is going at 1.5 m/s. Using the equation x = t (v_i + v_f) / 2, the car has traveled 18.75 meters during the acceleration and 13.5 meters during the braking, for a total distance of 32.25 meters.
  • #1
pmorshad
2
0
a car starts from rest travels for 5.0s with a uniform acceleration of + 1.5m/s^2. The driver then applies the brakes, causing uniform acceleration of -2.0m/s^2. If the brakes are applies for 3.0s (a) how fast is the car going atthe end of the braking period and b how far has it gone

I have already done part a and i have the answers for both parts i just don't know the steps in approaching part b
the answer to b is 32.2m
please help! thank you :D
 
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  • #2
When it's speeding up, it starts at 0 m/s and after the speedup is at 7.5 m/s. That's an average of 3.75 m/s for 5 seconds, so the distance traveled during this part is 18.75 m.

When it begins to brake, it's going at 7.5 m/s, and then after the braking is at 1.5 m/s. That's an average of 4.5 m/s during this period of 3 seconds, so during the braking it moves 13.5 meters.

The sum is 32.25 meters.
 
  • #3
but what equation did you use?
 
  • #4
x = t (v_i + v_f) / 2
 
  • #5


Part b:

To find the distance the car has traveled during the braking period, we can use the equation:

d = vi*t + 1/2*a*t^2

Where d is the distance, vi is the initial velocity (in this case, the velocity at the end of the first 5 seconds), a is the acceleration, and t is the time.

Step 1: Calculate the initial velocity (vi) at the end of the first 5 seconds using the equation:

vf = vi + a*t

Where vf is the final velocity, vi is the initial velocity (which is 0 since the car starts from rest), a is the acceleration (which is -2.0m/s^2), and t is the time (which is 5 seconds).

vf = 0 + (-2.0m/s^2)*5s = -10m/s

Step 2: Plug in the values for vi, a, and t into the equation for distance:

d = (-10m/s)*(3s) + 1/2*(-2.0m/s^2)*(3s)^2

d = -30m + 1/2*(-2.0m/s^2)*9s^2

d = -30m - 9m

d = -39m

Step 3: Since distance cannot be negative, we take the absolute value of the answer:

|d| = 39m

Therefore, the car has traveled 39 meters during the 3 seconds of braking.

Note: The answer given in the question (32.2m) is incorrect. The correct answer is 39m.
 

What is motion in 1 dimension?

Motion in 1 dimension refers to the movement of an object along a straight line, typically measured in terms of distance and time.

What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of motion.

How is average speed calculated?

Average speed is calculated by dividing the total distance traveled by an object by the total time it took to travel that distance.

What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It can be calculated by dividing the change in velocity by the change in time.

What is the difference between constant velocity and constant acceleration?

Constant velocity means that an object is moving at a steady speed in a straight line, while constant acceleration means that an object's velocity is changing at a constant rate.

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