a particle starts from rest at o and moves in a straight line so that t seconds after, its velocity vm/s is given by v=2t-t^2. given that particle comes instantanuously at rest at p. Find
i) time taken to reach p?
ii) the acceleration at p?
iii) distance OP?
iv)find the total distance travelled by the particle in the first 4 seconds of motion?
velocity = 2t-t^2
acceleration is dv/dt = 2-2t
distance is s = ∫v dt = t^2 - 1/3 t^3 + C
Since s(0) = 0, C=0
Now you can answer the questions
To find the answers to these questions, we need to integrate the equation for velocity with respect to time. Let's solve each question step by step:
i) To find the time taken to reach point P, we need to find the time when the velocity is zero.
Given: v = 2t - t^2
Setting v = 0, we can solve for t:
0 = 2t - t^2
Rearranging the equation, we get:
t^2 - 2t = 0
Factoring t, we have:
t(t - 2) = 0
So, t = 0 or t = 2.
The particle starts from rest at t = 0, so the time taken to reach point P is 2 seconds.
ii) To find the acceleration at point P, we can differentiate the equation for velocity with respect to time.
Given: v = 2t - t^2
Differentiating v with respect to t, we get:
a = d(2t - t^2)/dt
= 2 - 2t
Substitute t = 2 into the equation:
a(2) = 2 - 2(2)
= 2 - 4
= -2
So, the acceleration at point P is -2 m/s^2.
iii) To find the distance OP, we need to find the displacement. Displacement is the area under the velocity-time graph.
Given: v = 2t - t^2
Integrating v with respect to t, we get the displacement:
s = ∫(2t - t^2) dt
= t^2 - (t^3)/3 + C
To find the constant of integration (C), we use the initial condition that the particle starts from rest at t = 0. Therefore, s = 0 at t = 0.
0 = (0)^2 - (0^3)/3 + C
0 = 0 - 0 + C
C = 0
So, the displacement equation is:
s = t^2 - (t^3)/3
To find the distance OP, we substitute t = 2 into the equation:
s(2) = (2)^2 - ((2)^3)/3
= 4 - 8/3
= 4 - 2.67
= 1.33 meters
Therefore, the distance OP is 1.33 meters.
iv) To find the total distance traveled by the particle in the first 4 seconds of motion, we need to find the area under the velocity-time graph from t = 0 to t = 4.
Given: v = 2t - t^2
Integrating v with respect to t, we get the displacement:
s = ∫(2t - t^2) dt
= t^2 - (t^3)/3 + C
We already know C = 0 (from the previous calculation of displacement).
Substitute the limits t = 0 to t = 4 into the equation:
s(4) - s(0) = (4)^2 - ((4)^3)/3 - [0 - 0]
= 16 - (64)/3
= 16 - 21.33
= -5.33 meters
The total distance traveled by the particle in the first 4 seconds is 5.33 meters.
Note that the value is negative because the displacement equation gives a negative value in this case. However, distance does not have direction and is always positive, so the negative sign is ignored.
I hope this step-by-step explanation helps!
To find the answers to these questions, we need to analyze the given velocity function and use some basic principles of calculus. Let's go step by step:
i) To find the time taken to reach point P, we need to determine when the velocity is zero. We can do this by setting the velocity function v = 2t - t^2 equal to zero and solving for t:
2t - t^2 = 0
Factoring out t from the equation, we have:
t(2 - t) = 0
This equation is satisfied when either t = 0 or 2 - t = 0. We can ignore the t = 0 solution because the particle starts from rest at that point. Thus, solving 2 - t = 0, we find:
t = 2
Therefore, the time taken to reach point P is 2 seconds.
ii) The acceleration at point P can be found by differentiating the velocity function with respect to time. The derivative of v(t) = 2t - t^2 is the acceleration function a(t). Differentiating with respect to t, we get:
a(t) = d/dt(2t - t^2) = 2 - 2t
Substituting t = 2 (since we are interested in the acceleration at point P), we find:
a(2) = 2 - 2(2) = -2 m/s^2
Therefore, the acceleration at point P is -2 m/s^2.
iii) The distance OP can be found by integrating the velocity function from 0 to 2 seconds, which gives us the displacement over that time interval:
OP = ∫[0 to 2](2t - t^2) dt
Integrating term by term, we get:
OP = [t^2 - (t^3)/3] evaluated from 0 to 2
OP = (2^2 - (2^3)/3) - (0^2 - (0^3)/3)
OP = (4 - 8/3) - 0
OP = (12/3 - 8/3) = 4/3 m
Therefore, the distance OP is 4/3 meters.
iv) To find the total distance traveled by the particle in the first 4 seconds of motion, we need to calculate the displacement within that time interval. Since the particle changes direction at point P, we need to calculate the distance from O to P and back to O.
The distance from O to P is given by the magnitude of OP:
OP_distance = |OP| = |4/3| = 4/3 m
The distance from P to O is equal to |OP| since the magnitudes are the same.
Adding the distances together, we get the total distance traveled:
Total distance = OP_distance + PO_distance = (4/3) + (4/3) = 8/3 m
Therefore, the total distance traveled by the particle in the first 4 seconds of motion is 8/3 meters.