a stunt car driver testing the use of air bags drives a car at a constant speed of 25m/s for a total of 100 meters. he applies his brakes and accelerates uniformlu to a stop just as he reaches the wall 50 meters away. SKETCH THE A GRAPH SHOWING POSITION V.S TIME and VELOCITY v.s TIME

also how long does it take the car to travel the first 100 meters?

first do the stopping problem

d = Vi t + .5 a t^2
d = 25 t + .5 a t^2
50 = 25 t + .5 a t^2

v = Vi + a t
0 = 25 + at
a t = -25
a = -25/t
combine 50 = 25 t - .5(25t)
50 = .5 (25 t)
t = 4 seconds to stop after applying brakes
a = -25/4 while stopping
Now do the problem
for the first part
x = 25 t until t = 2 where x = 50
then x = 50 + 25(t-2) - (25/8) (t-2)^2
until stopped at x = 100 and t = 6
during deacceleration v = 25 -(25/4)(t-2)

To sketch the graphs showing position vs. time and velocity vs. time, we can use the following steps:

1. Determine the time it takes for the car to travel the first 100 meters.
2. Divide the total distance into two parts: the first 100 meters at a constant speed and the remaining 50 meters while accelerating to a stop.
3. Calculate the time it takes for the car to travel the remaining 50 meters.
4. Use the equations of motion to determine the position and velocity at any given time during each segment.
5. Plot the calculated values on the graph.

Let's go through these steps and calculate the time it takes for the car to travel the first 100 meters:

1. To find the time it takes for the car to travel the first 100 meters, we can use the equation: distance = velocity × time.
The distance traveled is 100 meters, and the speed is constant at 25 m/s.
So, we can rearrange the equation to solve for time:
time = distance / velocity = 100 m / 25 m/s = 4 seconds.

Now, let's divide the total distance into two parts:

- Part 1: The first 100 meters traveled at a constant speed of 25 m/s, which takes 4 seconds.
- Part 2: The remaining 50 meters while the car accelerates uniformly to a stop.

To calculate the time it takes for the car to travel the remaining 50 meters, we need to find its final velocity when it reaches the wall.

Since the car is decelerating uniformly, its final velocity (vf) is 0 m/s.

We can use the equation of motion: vf^2 = vi^2 + 2aΔx, where vf is the final velocity, vi is the initial velocity, a is acceleration, and Δx is the distance.

Rearranging this equation, we get: a = (vf^2 - vi^2) / (2Δx)

Substituting the given values, we have: a = (0^2 - 25^2) / (2 * 50) = -625 / 100 = -6.25 m/s^2.

Since acceleration (a = -6.25 m/s^2) is negative, it indicates deceleration.

Now, we can use the kinematic equation: vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is acceleration, and t is time.

Substituting the values, we have: 0 = 25 + (-6.25)t.

Solving for t, we get: t = 25 / 6.25 = 4 seconds.

We now have the time it takes for the car to travel the remaining 50 meters, which is also 4 seconds.

Now that we have all the necessary values, we can plot the position vs. time and velocity vs. time graphs.

Position vs. time:

- During the first 4 seconds (100 meters), the position remains constant at 100 meters.
- During the next 4 seconds (50 meters), the position decreases linearly from 100 meters to 50 meters.

Velocity vs. time:

- During the first 4 seconds (100 meters), the velocity remains constant at 25 m/s.
- During the next 4 seconds (50 meters), the velocity decreases linearly from 25 m/s to 0 m/s.

Now, using these plotted values, you can sketch the graphs showing position vs. time and velocity vs. time.

To sketch the graphs showing position vs. time and velocity vs. time, we will break the car's motion into three stages:

Stage 1: Constant Speed:
During this stage, the car moves at a constant speed of 25 m/s for a total of 100 meters.

Stage 2: Braking:
The driver applies the brakes and begins to decelerate uniformly until he reaches the wall.

Stage 3: Stopped:
The car comes to a stop just as it reaches the wall.

Here is a step-by-step guide on how to sketch the graphs:

1. Position vs. Time Graph:
- For the first 100 meters, the car moves at a constant speed. Therefore, the position vs. time graph will be a straight, horizontal line at a height of 100 meters for a duration of time.

2. Velocity vs. Time Graph:
- Since the car is moving at a constant speed of 25 m/s, the velocity vs. time graph will be a straight horizontal line at a height of 25 m/s for the duration the car travels the first 100 meters.

3. Braking Stage:
- Once the car reaches the 100-meter mark, the driver applies the brakes and decelerates uniformly.
- The velocity decreases steadily until the car comes to a stop.
- The velocity vs. time graph will show a negative slope, gradually decreasing until the velocity reaches zero.

4. Stopped Stage:
- After the braking stage, the car comes to a complete stop just as it reaches the wall.
- Both the position and velocity will remain constant at zero during this stage.

As for the time it takes for the car to travel the first 100 meters, we need to determine the duration of stage 1.
The time can be calculated using the formula:
Time = distance / speed

Time = 100 meters / 25 m/s
Time = 4 seconds

Therefore, it takes the car 4 seconds to travel the first 100 meters.