A car A is 300m behind a car B and both are travelling in the same direction at 25m/s. Both cars begin at the same moment to accelerate at steady rate, but A has a greater acceleration and passes B in 20sec by which time B is doing 27•5m/s. Neglecting the length of the cars, calculate the acceleration of A, it's speed at the moment of passing B and the distance it has travelled in the 20sec.
V = Vo + a*t = 27.5, 25 + a*20 = 27.5, 20a = 2.5, a = 0.125 m/s^2. = Acceleration of car B.
Vo*t + 0.5a*t^2 = Vo*t + o.5a*t^2 + 300,
Vo*t-Vo*t + 0.5a*t^2 = 0.5*0.125*20^2 + 300,
0.5a*20^2 = 25 + 300 = 325,
200a = 325, a = 1.625 = Acceleration of car A.
Va = Vo + a*t = 25 + 1.625*20 = 57.5 m/s. = Speed of car A at the moment of passing B.
d = 25*20 + 1.625*20^2 = 1150 m.
=
To solve this problem, we need to apply the kinematic equations of motion. We'll break down the problem into smaller steps to find the required values.
Step 1: Find the initial speed of car B
Given:
- Car B's speed = 27.5 m/s
- Time taken = 20 s
The initial speed of car B is the speed at which it started accelerating. Since the problem states that both cars started accelerating simultaneously, their initial speeds should be the same. Therefore, the initial speed of car B is also 25 m/s.
Step 2: Find the acceleration of car B
Given:
- Initial speed of car B = 25 m/s
- Final speed of car B = 27.5 m/s
- Time taken = 20 s
We will use the formula:
acceleration (a) = (final velocity - initial velocity) / time
Substituting the given values, we have:
acceleration of car B (ab) = (27.5 - 25) / 20
Step 3: Find the speed of car A at the moment of passing car B
Given:
- Final speed of car B = 27.5 m/s
- Time taken = 20 s
We can assume that car A had a constant acceleration throughout the 20 seconds. This implies that its final speed is the difference between the speeds of car A and car B when they passed.
Let's assume the final speed of car A at the moment of passing car B is v.
v = final speed of car A - final speed of car B
v = v - 27.5
Since car A started with a higher acceleration, it would have overtaken car B. Therefore, at the moment of passing, car A's final speed is equal to the final speed of car B.
v = 27.5 m/s
The speed of car A at the moment of passing car B is 27.5 m/s.
Step 4: Find the acceleration of car A
Given:
- Time taken = 20 s
- Speed of car A at the moment of passing car B = 27.5 m/s
We can use the formula we derived earlier to find the acceleration of car A:
acceleration of car A (aa) = (final velocity of car A - initial velocity of car A) / time
Rearranging the formula to solve for acceleration of car A:
acceleration of car A = (v - 25) / 20
acceleration of car A = (27.5 - 25) / 20
Step 5: Find the distance traveled by car A in 20 seconds
We can use the formula:
distance (d) = initial velocity (u) * time (t) + 0.5 * acceleration (a) * t^2
The initial velocity of car A is 0 because it started from rest. The time taken is 20 seconds, and we have already calculated the acceleration of car A.
distance traveled by car A in 20 seconds = 0.5 * acceleration of car A * (time^2)
distance traveled by car A in 20 seconds = 0.5 * (27.5 - 25) / 20 * (20^2)
Now, substitute the values into the equation to find the distance traveled by car A.