A car starts from rest and accelerates at a rate of 5.75 m/s/s for a time (t1) of 6.76 seconds. The car then maintains a constant speed for a time of 7.32 seconds (t2 minus t1). Finally, the car slows down at a rate of -4.21 m/s/s.
for each segment of the trip,
s = vt + 1/2 at^2
so plug in your numbers.
Determine the total distance traveled by the car from its starting position to its stopping position.
Huh. I can see I need to get you started.
so, for the first part of the trip,
s = 0 + 5.75/2 * 6.76^2 = 131.38 m
Now do the 2nd part and the 3rd part, and add them up.
I got the second part, 131.38 and 284.53. How do I get the time for the 3rd part?
To find the total distance traveled by the car, we need to break down the problem into three parts: the initial acceleration, the constant speed, and the final deceleration.
1. Initial Acceleration:
The car starts from rest and accelerates at a rate of 5.75 m/s/s for a time of 6.76 seconds (t1). To find the distance covered during this period, we can use the kinematic equation:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
Since the car starts from rest (initial velocity = 0), the equation simplifies to:
Distance1 = 0.5 * acceleration * time^2
Substituting the given values:
Distance1 = 0.5 * 5.75 m/s^2 * (6.76 s)^2
Distance1 ≈ 117.48 meters
2. Constant Speed:
The car maintains a constant speed for a time of 7.32 seconds (t2 minus t1). When the speed is constant, acceleration is zero. Therefore, there is no distance traveled during this period.
Distance2 = 0 meters
3. Final Deceleration:
The car slows down at a rate of -4.21 m/s/s. Since the acceleration is negative, the final deceleration is considered a negative acceleration. We can use the same kinematic equation as before to find the distance covered during this period:
Distance3 = (Initial velocity * time) + (0.5 * acceleration * time^2)
Since the initial velocity at the end of the constant speed period is the same as the final velocity of the previous period, we can use the formula:
Distance3 = (0 * t) + (0.5 * acceleration * t^2)
Substituting the given values:
Distance3 = 0.5 * (-4.21 m/s^2) * (t)^2
To find the value of 't', we need to subtract the total time of the previous periods (t1 + t2) from the total time given (t1 + t2 + t3):
t = t3 = (t1 + t2 + t3) - (t1 + t2) = t3 - t2
Substituting the values:
t ≈ 7.32 s - 6.76 s
t ≈ 0.56 s
Now substituting 't' back into the distance equation:
Distance3 = 0.5 * (-4.21 m/s^2) * (0.56 s)^2
Distance3 ≈ -0.83 meters (negative sign indicates deceleration)
Total distance traveled = Distance1 + Distance2 + Distance3
Total distance traveled ≈ 117.48 m + 0 m - 0.83 m
Total distance traveled ≈ 116.65 meters
Therefore, the car covers a total distance of approximately 116.65 meters.