on the same Lane of a road a car A is following car B at a center distance of 25 m. both cars are traveling at a speed of 100 kmph when car A attempts to overtake car B at a uniform acceleration of 0.8 m/s^2.after traveling for 5 sec at that acceleration rate to overtake car B and comes to a position 20m in front of car B in another 5 sec. what is the increased uniform acceleration rate?
1.2 m/sec²
To find the increased uniform acceleration rate, we'll need to break down the problem step-by-step. Here's how we can calculate it:
Step 1: Convert the speeds to m/s.
Given:
Speed of car A = 100 km/h
Speed of car B = 100 km/h
To convert km/h to m/s, divide by 3.6:
Speed of car A = (100 km/h) / (3.6 m/s) = (100 × 1000 m) / (3.6 × 3600 s) = 27.78 m/s
Speed of car B = 27.78 m/s
Step 2: Calculate the initial relative velocity of car A with respect to car B.
Relative velocity = Speed of car A - Speed of car B
Relative velocity = 27.78 m/s - 27.78 m/s = 0 m/s
Step 3: Determine the time it takes for car A to overtake car B.
Given:
Acceleration of car A = 0.8 m/s^2
Initial relative velocity = 0 m/s
We can use the formula: Distance = (Initial velocity × Time) + (0.5 × Acceleration × Time^2)
Since the distance traveled by car A is 25 m and the relative velocity is 0 m/s, we have:
25 m = (0 m/s × Time) + (0.5 × 0.8 m/s^2 × Time^2)
Simplifying the equation:
25 m = 0.4 m/s^2 × Time^2
Step 4: Solve for the time it takes for car A to overtake car B.
Rearranging the equation:
Time^2 = (25 m) / (0.4 m/s^2)
Time^2 = 62.5 s^2
Taking the square root of both sides:
Time = √(62.5 s^2) ≈ 7.91 s
Step 5: Calculate the final relative velocity of car A with respect to car B.
Given:
Time = 5 s
We can use the formula: Final velocity = Initial velocity + (Acceleration × Time)
Since the initial relative velocity is 0 m/s and the time is 5 s, we have:
Final relative velocity = 0 m/s + (0.8 m/s^2 × 5 s)
Final relative velocity = 4 m/s
Step 6: Determine the distance covered by car A to overtake car B.
Given:
Final relative velocity = 4 m/s
Time = 5 s
We can use the formula: Distance = (Initial velocity × Time) + (0.5 × Acceleration × Time^2)
Since the initial relative velocity is 0 m/s and the time is 5 s, we have:
Distance = (0 m/s × 5 s) + (0.5 × Acceleration × (5 s)^2)
Distance = 0 + (0.5 × Acceleration × 25 s^2)
Distance = 12.5 × Acceleration
Step 7: Solve for the distance covered by car A.
Since we know that the distance covered by car A is 20 m, we can set up the equation:
20 m = 12.5 × Acceleration
Step 8: Solve for the increased uniform acceleration rate.
Rearranging the equation:
Acceleration = (20 m) / 12.5
Acceleration = 1.6 m/s^2
Therefore, the increased uniform acceleration rate is 1.6 m/s^2.
To find the increased uniform acceleration rate, we can use the formula:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
First, let's convert the speed from km/h to m/s:
100 km/h = (100 * 1000) m/3600 s = 27.78 m/s
Now, let's analyze the movements of car A. It begins with an initial velocity (u) of 27.78 m/s and accelerates at a rate of 0.8 m/s^2 for 5 seconds. We can use the first equation to find the final velocity (v1) after 5 seconds:
v1 = 27.78 + (0.8 * 5)
v1 = 31.78 m/s
Next, we need to determine the position of car A after 5 seconds. We can use the formula:
s = ut + 0.5at^2
Where:
s = displacement
u = initial velocity
t = time
a = acceleration
Using the values we have:
s = (27.78 * 5) + (0.5 * 0.8 * 5^2)
s = 138.9 + 10
s = 148.9 m
So after 5 seconds, car A has traveled a distance of 148.9 m. Given that it is 20 m ahead of car B, the initial distance between the two cars (D) can be calculated as follows:
D = 148.9 - 20
D = 128.9 m
Now, let's analyze the next phase. Car A continues to accelerate for another 5 seconds until it is 20 m ahead of car B. We can use the formula again to find the final velocity (v2) after these additional 5 seconds:
v2 = 31.78 + (a * 5)
v2 = 31.78 + (a * 5)
The total distance traveled by car A is given by:
s = ut + 0.5at^2
Using the values we have:
s = (31.78 * 5) + (0.5a * 5^2)
s = 158.9 + 12.5a
Since the total distance traveled in the second phase is 128.9 m (20 m less than the total distance in the first phase), we can set up the equation:
158.9 + 12.5a = 128.9
Rearranging the equation:
12.5a = 128.9 - 158.9
12.5a = -30
Finally, solving for acceleration (a):
a = -30 / 12.5
a ≈ -2.4 m/s^2
Therefore, the increased uniform acceleration rate is approximately -2.4 m/s^2. The negative sign indicates that car A is decelerating during the second phase.