1. a car traveling on a dry road with a velocity of +32.0m/s. The driver slams on the breaks and skids to a halt with an acceleration of -8.00m/s^2. on an icy road the car would have skidded to a halt with an acceleration of -3.00m/s^2. how much further would the car have skidded on the icy road compared to the dry road?
2. A jogger accelerates from rest to 3.0m/s in 2.0s. A car accelerates from 38.0 to 41.0 m/s also in 2.0s. Find the acceleration of the jogger and determine the acceleration of the car. Does the car travel further than the jogger during the 2.0s? if so how much further?
A jogger accelerates from rest to 3.1 m/s in 2.2 s. A car accelerates from 39.0 to 40.0 m/s also in 2.2 s. Find the acceleration of the jogger and determine the acceleration of the car. Does the car travel further than the jogger during the 2.2s? if so how much further?
Car on an icy road would have skidded 107 meters more than on a dry road.
1. To find how much further the car would have skidded on the icy road compared to the dry road, we need to find the distance traveled on each road.
First, let's find the distance traveled on the dry road. We can use the equation:
vf^2 = vi^2 + 2ad
where vf is the final velocity (0 m/s since the car comes to a halt), vi is the initial velocity (+32.0 m/s), a is the acceleration (-8.00 m/s^2), and d is the distance traveled.
Rearranging the equation, we get:
d = (vf^2 - vi^2) / (2a)
Plugging in the values, we get:
d = (0^2 - (+32.0)^2) / (2*(-8.00))
= (-1024.0) / (-16.0)
= 64.0 m
Therefore, the car travels 64.0 meters on the dry road before coming to a halt.
Now, let's find the distance traveled on the icy road. Using the same equation:
d = (vf^2 - vi^2) / (2a)
But with the acceleration on the icy road (-3.00 m/s^2), we get:
d = (0^2 - (+32.0)^2) / (2*(-3.00))
= (-1024.0) / (-6.00)
= 170.67 m (rounded to two decimal places)
Therefore, the car would have skidded approximately 170.67 meters on the icy road, compared to 64.0 meters on the dry road. The car would have skidded approximately 106.67 meters further on the icy road.
2. To find the acceleration of the jogger and the car, we can use the equation:
a = (vf - vi) / t
where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken.
For the jogger:
vi = 0 m/s (since the jogger starts from rest)
vf = 3.0 m/s
t = 2.0 s
Using the equation, we get:
a = (3.0 - 0) / 2.0
= 1.5 m/s^2
Therefore, the acceleration of the jogger is 1.5 m/s^2.
For the car:
vi = 38.0 m/s
vf = 41.0 m/s
t = 2.0 s
Using the equation, we get:
a = (41.0 - 38.0) / 2.0
= 1.5 m/s^2
Therefore, the acceleration of the car is also 1.5 m/s^2.
To determine if the car travels further than the jogger during the 2.0 seconds, we need to find the distance traveled by each.
For the jogger:
Using the equation:
d = vi * t + (1/2) * a * t^2
where d is the distance, vi is the initial velocity, t is the time taken, and a is the acceleration, we get:
d = 0 * 2.0 + (1/2) * (1.5) * (2.0)^2
= 0 + 1.5 * 4.0
= 6.0 m
Therefore, the jogger travels 6.0 meters during the 2.0 seconds.
For the car:
Using the same equation, but with different values:
d = vi * t + (1/2) * a * t^2
where vi = 38.0 m/s, t = 2.0 s, and a = 1.5 m/s^2, we get:
d = 38.0 * 2.0 + (1/2) * (1.5) * (2.0)^2
= 76.0 + 1.5 * 4.0
= 76.0 + 6.0
= 82.0 m
Therefore, the car travels 82.0 meters during the 2.0 seconds.
The car travels further than the jogger during the 2.0 seconds, by a distance of 82.0 - 6.0 = 76.0 meters.
To answer both of these questions, we will use the kinematic equations of motion, specifically the equations of motion under constant acceleration.
1. To find how much further the car would have skidded on the icy road compared to the dry road, we know the initial velocity (v0), the final velocity (vf), and the acceleration (a) for both scenarios. The distance (d) can be calculated using the equation:
d = (vf^2 - v0^2) / (2a)
For the dry road:
v0 = 32.0 m/s
vf = 0 m/s
a = -8.00 m/s^2
Plugging these values into the equation, we get:
d1 = (0^2 - 32.0^2) / (2 * -8.00)
Similarly, for the icy road:
v0 = 32.0 m/s
vf = 0 m/s
a = -3.00 m/s^2
Plugging these values into the equation, we get:
d2 = (0^2 - 32.0^2) / (2 * -3.00)
To find the difference in skidding distances between the two roads, we subtract d2 from d1:
Δd = d1 - d2
2. To find the accelerations of the jogger and the car during the given time interval, we will use the equation of motion:
vf = v0 + at
For the jogger:
v0 = 0 m/s
vf = 3.0 m/s
t = 2.0 s
Plugging these values into the equation, we can solve for a:
3.0 = 0 + a * 2.0
For the car:
v0 = 38.0 m/s
vf = 41.0 m/s
t = 2.0 s
Again, plugging these values into the equation, we can solve for a:
41.0 = 38.0 + a * 2.0
Comparing the distances traveled during the 2.0 seconds for the jogger and the car, we use the equation:
d = v0 * t + 0.5 * a * t^2
For the jogger:
v0 = 0 m/s
a (from previous calculation)
t = 2.0 s
Plugging these values into the equation, we can solve for d1.
For the car:
v0 = 38.0 m/s
a (from previous calculation)
t = 2.0 s
Plugging these values into the equation, we can solve for d2.
To find the difference in distances traveled, we subtract d1 from d2:
Δd = d2 - d1