Last question from me
Suppose the driver in this example is now moving with speed 39.6 m/s, and slams on the brakes, stopping the car in 4.1 s.
(a) Find the acceleration assuming the acceleration is constant.
m/s2
(b) Find the distance the car travels, assuming the acceleration is constant.
m
(c) Find the average velocity.
m/s
average speed during stop = (39.6+0)/2 = 19.8 m/s (a is constant so v is linear) That is part C answer.
time to stop = 4.1 s
so distance to stop = 4.1*19.8 = 81.2 meters (part B answer)
v = Vi + a t
0 = 39.6 + a(4.1)
a = -9.66 m/s^2 part A answer
a is negative of course because de accelerating
By the way, that is a little less than 1 g. He indeed slammed the brake on.
a. Vo + a*t = 0.
30.6 + a*4.1 = 0,
a = -7.5 m/s^2.
b. Vo^2 + 2a*d = V^2.
30.6^2 - 7.5d = 0,
d =
c. Vavg. = (Vo = +V)/2 = (30.6+0)/2 =
39.6 I think :)
Yes, 39.6. My mistake.
help
ppose the driver in this example is now moving with speed 38.9 m/s, and slams on the brakes, stopping the car in 4.4 s.
(a) Find the acceleration assuming the acceleration is constant.
Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2
(b) Find the distance the car travels, assuming the acceleration is constant.
Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 10%. Double check your calculations. m
(c) Find the average velocity.
Incorrect: Your answer is incorrect.
m/s
To find the answers to these questions, we can use the equations of motion. Here's how we can solve each part:
(a) Find the acceleration assuming it is constant:
We know the initial velocity, final velocity, and time taken to stop. The equation that relates these variables is:
v = u + at
Where:
v = final velocity (0 m/s because the car stops)
u = initial velocity (39.6 m/s)
a = acceleration (to be determined)
t = time taken to stop (4.1 s)
Rearranging the equation to solve for acceleration, we have:
a = (v - u) / t
Plugging in the values:
a = (0 m/s - 39.6 m/s) / 4.1 s
Evaluating this expression gives us the value of acceleration in m/s^2.
(b) Find the distance the car travels assuming the acceleration is constant:
To find the distance traveled, we can use the equation:
s = ut + (1/2)at^2
Where:
s = distance traveled (to be determined)
u = initial velocity (39.6 m/s)
a = acceleration (to be determined)
t = time taken to stop (4.1 s)
Plugging in the values:
s = (39.6 m/s)(4.1 s) + (1/2)(acceleration)(4.1 s)^2
We can substitute the value of acceleration from part (a) into this equation to find the distance traveled.
(c) Find the average velocity:
The average velocity can be calculated using the equation:
average velocity (v_avg) = total displacement / total time
Since the car stops, the displacement is equal to the distance traveled (s) in this case. Plug in the values for distance and time taken to stop to find the average velocity.
By following these steps and plugging in the appropriate values, you should be able to find the acceleration, distance traveled, and average velocity in this scenario.