A car is traveling along a straight road at a velocity of +27.7 m/s when its engine cuts out. For the next 1.94 seconds, the car slows down, and its average acceleration is . For the next 4.92 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 6.86-second period is +20.3 m/s. The ratio of the average acceleration values is = 1.35. Find the velocity of the car at the end of the initial 1.94-second interval.
To find the velocity of the car at the end of the initial 1.94-second interval, we can use the relationship between acceleration, velocity, and time, which can be represented by the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = average acceleration
t = time
Let's define the variables given in the problem:
u = 27.7 m/s (initial velocity)
a1 = average acceleration for the first 1.94 seconds
a2 = average acceleration for the next 4.92 seconds
v = 20.3 m/s (final velocity)
t1 = 1.94 seconds (time for the first interval)
From the given information, we know that:
a2/a1 = 1.35
We also know that the car is slowing down, so the acceleration values will be negative.
Now, let's calculate the unknown variable using the given information:
1. Calculate the average acceleration for the second interval:
a2 = 1.35 * a1
2. Calculate the time for the second interval:
t2 = 6.86 seconds - 1.94 seconds = 4.92 seconds
3. Calculate the change in velocity during the first interval:
∆v1 = v - u
4. Use the equation v = u + at to find the velocity at the end of the first interval:
v1 = u + (a1 * t1)
5. Use the equation v = u + at to find the velocity at the end of the second interval:
v2 = v1 + (a2 * t2)
6. Use the equation ∆v = at to find the change in velocity during the second interval:
∆v2 = a2 * t2
7. Calculate the final velocity using the change in velocity values:
v = v1 + ∆v1 + ∆v2
Plug in the known values into the above equations to find the velocity of the car at the end of the initial 1.94-second interval.
To find the velocity of the car at the end of the initial 1.94-second interval, we can use the equations of motion.
Let's denote the initial velocity of the car as 'v0', the final velocity of the car at the end of the 6.86-second period as 'vf', the time of the initial interval as 't1', the time of the second interval as 't2', and the average accelerations as 'a1' and 'a2' respectively.
Given:
Initial velocity (v0) = +27.7 m/s
Time of the initial interval (t1) = 1.94 seconds
First average acceleration (a1) = ? (to be calculated)
Time of the second interval (t2) = 4.92 seconds
Second average acceleration (a2) = ? (to be calculated)
Final velocity (vf) = +20.3 m/s
We can calculate the first average acceleration (a1) using the formula:
a1 = (vf - v0) / t1
Substituting the values in the formula, we get:
a1 = (20.3 - 27.7) / 1.94
Simplifying, we have:
a1 = -7.4 / 1.94
Next, we can calculate the second average acceleration (a2) using the formula:
a2 = 1.35 * a1
Substituting the value of a1, we get:
a2 = 1.35 * (-7.4 / 1.94)
Now, to find the velocity of the car at the end of the initial 1.94-second interval, we can use the formula:
vf = v0 + a1 * t1
Rearranging the formula, we have:
vf - v0 = a1 * t1
Substituting the values, we get:
20.3 - 27.7 = (-7.4 / 1.94) * 1.94
Simplifying, we have:
-7.4 = -7.4
Therefore, the velocity of the car at the end of the initial 1.94-second interval is -7.4 m/s.