A dragster accelerates from rest for a distance of 450m at 14 m/s^2. A parachute is then used to slow it down to a stop if the parachute gives the dragster an acceleration of -7m/s^2 how far has the dragster traveled before stopping?
Well, aren't dragsters just fancy racecars with some extra flair? Now, let's calculate the distance this dragster has traveled before coming to a stop.
First, we need to find out how long it took for the dragster to accelerate. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity (u) is 0 m/s (since it starts from rest) and the acceleration (a) is 14 m/s^2, we can rearrange the equation to solve for time (t). So, t = (v - u) / a.
Plugging in the values, we get t = (14 m/s - 0 m/s) / 14 m/s^2 = 1 second.
Now, let's find out how far the dragster traveled during this period of acceleration. We can use the formula s = ut + 1/2at^2, where s is the distance.
Given that the initial velocity (u) is 0 m/s, the acceleration (a) is 14 m/s^2, and the time (t) is 1 second, we can calculate the distance (s). So, s = 0 m/s * 1 s + 1/2 * 14 m/s^2 * (1 s)^2 = 7 meters.
Now, the dragster needs to come to a stop using a parachute. The acceleration caused by the parachute (a) is -7 m/s^2, which means it's in the opposite direction.
To calculate the distance covered during deceleration, we can use the formula v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Given that the final velocity (v) is 0 m/s (since it comes to a stop), the initial velocity (u) is 14 m/s, and the acceleration (a) is -7 m/s^2, we can rearrange the equation to solve for distance (s). So, s = (v^2 - u^2) / (2a).
Plugging in the values, we get s = (0 m/s)^2 - (14 m/s)^2 / (2 * -7 m/s^2) = 98 meters.
Now, to find the total distance covered by the dragster, we add the distance during acceleration (7 meters) to the distance during deceleration (98 meters).
Therefore, the dragster has traveled a total distance of 7 meters + 98 meters = 105 meters before coming to a stop.
To find the distance traveled before stopping, we need to calculate the distance traveled during acceleration and the distance traveled during deceleration.
First, let's calculate the distance traveled during acceleration:
We are given:
Initial velocity (u) = 0 m/s (rest)
Acceleration (a1) = 14 m/s^2
Distance (s1) = 450 m
We can use the formula:
s1 = (u * t) + (0.5 * a1 * t^2)
Since the initial velocity is 0, the formula becomes:
s1 = 0.5 * a1 * t^2
Rearranging the formula, we get:
t^2 = (2 * s1) / a1
Substituting the given values, we get:
t^2 = (2 * 450) / 14
Calculating, we find:
t^2 = 900 / 14
t^2 ≈ 64.2857
Taking the square root of both sides, we get:
t ≈ 8.0288 seconds
Now, we need to calculate the distance traveled during deceleration:
We are given:
Initial velocity (u) = 0 m/s (rest)
Acceleration (a2) = -7 m/s^2
Using the formula:
s2 = (u * t) + (0.5 * a2 * t^2)
Since the initial velocity is 0, the formula becomes:
s2 = 0.5 * a2 * t^2
Substituting the given values, we get:
s2 = 0.5 * (-7) * t^2
s2 = -3.5 * t^2
Calculating, we find:
s2 = -3.5 * (8.0288)^2
s2 ≈ -3.5 * 64.4623
s2 ≈ -225.6181m
Note: The negative sign indicates the direction of deceleration.
Now, let's add the distances traveled during acceleration and deceleration:
Total distance = s1 + s2
Total distance = 450m - 225.6181m
Total distance ≈ 224.3819m
Thus, the dragster traveled approximately 224.3819 meters before coming to a stop.
To find the distance traveled by the dragster before stopping, we need to calculate the distance covered during the acceleration phase and the distance covered during the deceleration phase.
Let's start with the acceleration phase. The formula to calculate distance using constant acceleration is:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
In this case, the initial velocity is 0 m/s (since the dragster starts from rest), the acceleration is 14 m/s^2, and we need to find the time it takes to cover the distance of 450m.
Using the formula above, we rearrange it to solve for time:
distance = (0 * time) + (0.5 * 14 * time^2)
450 = 0.5 * 14 * time^2
Now, let's solve for time:
450 = 7 * time^2
time^2 = 450 / 7
time^2 ≈ 64.29
time ≈ √64.29
time ≈ 8 seconds
So, the dragster takes 8 seconds to cover the distance of 450m during the acceleration phase.
Now, we move to the deceleration phase. We have an initial velocity of 14 m/s (from the acceleration phase), the acceleration is -7 m/s^2 (opposite direction, since the dragster is slowing down), and we need to find the distance.
Using the same formula as before:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Here, the initial velocity is 14 m/s, the acceleration is -7 m/s^2, and the time is equal to the time taken during the acceleration phase, which is 8 seconds.
Plugging in the values, we have:
distance = (14 * 8) + (0.5 * -7 * 8^2)
distance = 112 - 224
distance = -112
The negative sign indicates that the dragster covers a negative distance, which means it is moving in the opposite direction.
Therefore, the dragster has traveled a distance of 112 meters in the opposite direction before coming to a stop.
The distance travelled while accelerating at a rate a=14 m/s^2 is X = 450 m and the maximum velocity attained at that time is V = sqrt(2aX) = 112.2 m/s. The time required to decelerate to a stop at a rate a' = 7 m/s^2 is t = V/a' = 16.0 s
The average velocity while decelerating is
v = V/2 = 56.1 m/s and the distance travelled while declerating is v*t
X' = 56.1 m/s*16 s = 900 m
The total distance travelled when it has stopped is 450 + 900 = 1350 m
Here is an easier way: Since the rate of deceleration is half the rate of acceleration, it takes twice as long to decelerate and one travels twice as far (900 m)while decereating.