A particle P, starting from rest at A, moves in a straight line ABCD. It accelerates uniformly at 5ms-2 from A to B. From B to C it travels at constant velocity, and from C to D it slows down uniformly at 4ms-2, coming to rest at D. Given that AB=10m and that the total time P the particle is in motion is 20s, find the distance BC
Let the three time periods be x,y,z seconds
x+y+z = 20
(5/2 x^2) + (5xy) + (5xz-2z^2) = 20
Now, since it decelerates at 4/5 the rate that it took to accelerate, it will take 5/4 as long to come to rest. So, our third equation is
z = 5/4 x
Now just solve for the time periods, and then you can get the three distances.
To find the distance BC, we need to first calculate the time it takes for the particle to travel from A to B.
The formula to calculate the time taken to travel a distance with uniform acceleration is given by:
t = (v - u) / a
Where,
t = time taken
v = final velocity
u = initial velocity
a = acceleration
In this case, the particle starts from rest, so the initial velocity u = 0 m/s. The final velocity v can be calculated using the equation:
v^2 = u^2 + 2as
Where,
s = distance
Given that AB = 10 m and a = 5 m/s^2, we can calculate the final velocity v as follows:
v^2 = 0^2 + 2 * 5 * 10
v^2 = 0 + 100
v^2 = 100
Taking the square root of both sides, we get:
v = 10 m/s
Now, we can substitute the values of v, u, and a into the formula for time to calculate the time taken to travel from A to B:
t1 = (v - u) / a
t1 = (10 - 0) / 5
t1 = 10 / 5
t1 = 2 s
Next, we need to calculate the time taken to travel from C to D. We are given that the particle decelerates uniformly at 4 m/s^2, and comes to rest at D. Similar to the previous calculation, we can use the formula for time:
t3 = (v - u) / a
In this case, the initial velocity u = 0 m/s (particle starts from rest), and the final velocity v = 0 m/s (particle comes to rest at D). Substituting these values into the formula, we have:
t3 = (0 - u) / a
t3 = (0 - 0) / 4
t3 = 0 / 4
t3 = 0 s
Now, we can calculate the time taken for the particle to travel from B to C:
Total time taken = Time taken from A to B + Time taken from B to C + Time taken from C to D
20 = t1 + t2 + t3
20 = 2 + t2 + 0
t2 = 20 - 2
t2 = 18 s
Since the particle travels at a constant velocity from B to C, the distance BC can be calculated using the following formula:
Distance = Velocity * Time
We are given that the total time taken from B to C is 18 s. Therefore:
BC = v * t2
BC = 10 * 18
BC = 180 m
Hence, the distance BC is 180 meters.
To find the distance BC, we need to determine the time it takes for the particle to travel from A to B and from C to D.
1. From A to B:
We know that the particle accelerates uniformly at 5 m/s^2. We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
The initial velocity, u, is 0 m/s since the particle starts from rest. The final velocity, v, can be found using the equation: v^2 = u^2 + 2as, where s is the distance traveled.
Rearranging the equation, we have v = √(u^2 + 2as).
Given that AB = 10 m and the particle starts from rest, the distance traveled, s, is 10 m.
Plugging in the values into the equation, we get v = √(0^2 + 2 * 5 * 10) = √100 = 10 m/s.
Next, we can find the time taken, t1, using the equation: t1 = (v - u) / a = (10 - 0) / 5 = 2 s.
2. From C to D:
The particle slows down uniformly at 4 m/s^2. Using the same equation of motion, v = u + at, we know that the final velocity, v, is 0 m/s since the particle comes to rest at D and the initial velocity, u, can be found using the equation: v^2 = u^2 + 2as.
Given that the distance traveled, s, is BC, we don't know the value yet, and the acceleration, a, is -4 m/s^2 (negative because the particle is slowing down).
Plugging in these values, we get 0 = u^2 + 2 * -4 * BC.
Simplifying the equation, we have u^2 = 8 * BC.
Since the velocity, u, is the same as the final velocity at B (10 m/s), we can substitute the value: (10)^2 = 8 * BC.
Solving for BC, we get BC = (10)^2 / 8 = 12.5 m.
Now, we know that the total time taken for the particle to move from A to B, stay at constant velocity from B to C, and then move from C to D is 20 s. We can use this information to find the time taken from B to C, t2.
The time taken for each leg of the journey can be represented as: t1 + t2 + t3 = 20.
From our previous calculations, we know t1 = 2 s. We also know that the particle travels at constant velocity from B to C, so the time taken, t2, will be (BC / constant velocity).
Plugging in the values, the equation becomes: 2 + (12.5 / constant velocity) + t3 = 20.
Since the particle travels at a constant velocity from B to C, the time taken, t2, is BC / constant velocity.
Substituting the values into the equation, we have: 2 + (12.5 / constant velocity) + t3 = 20.
Simplifying the equation, we get: (12.5 / constant velocity) + t3 = 18.
At this point, we need to know the value of the constant velocity to solve for t2. Unfortunately, the problem statement does not provide information about the constant velocity. Without that information, we cannot determine the exact value of BC.
Therefore, the distance BC cannot be determined without the value of the constant velocity from B to C.