a particle moves with acceleration a(t)=(t-2)m/s^2 along an x axis and has velocity v=(0)m/s at time t=0.find the displacement and distance travelled by the particle during the given time intervals 1is less than equal to t less than equal to 5
To find the displacement and distance travelled by the particle during the given time interval, we first need to find the velocity function of the particle.
The acceleration of the particle is given as a(t) = (t-2) m/s^2. We can find the velocity function by integrating the acceleration function with respect to time:
v(t) = ∫ a(t) dt
Integrating (t-2) with respect to t gives us:
v(t) = (1/2)t^2 - 2t + C
Using the given information that the velocity of the particle is 0 m/s at time t=0, we can plug in these values to solve for C:
v(0) = (1/2)(0)^2 - 2(0) + C
0 = 0 - 0 + C
C = 0
Therefore, the velocity function of the particle is:
v(t) = (1/2)t^2 - 2t
To find the displacement and distance travelled during the time interval 1 ≤ t ≤ 5, we need to calculate the definite integral of the absolute value of the velocity function within that interval.
Displacement:
The displacement of the particle is given by the integral of the velocity function:
Displacement = ∫[1, 5] |v(t)| dt
Since the absolute value function removes any negative sign, we can split up the integral into two parts:
Displacement = ∫[1, 5] v(t) dt - ∫[1, 5] (-v(t)) dt
Displacement = ∫[1, 5] [(1/2)t^2 - 2t] dt + ∫[1, 5] [(1/2)t^2 - 2t] dt
To find the integral of each term, we can use the power rule for integration:
Displacement = [(1/6)t^3 - t^2] evaluated from 1 to 5 + [(-1/2)t^2 + 2t] evaluated from 1 to 5
Displacement = [(1/6)(5^3) - (5^2)] - [(1/6)(1^3) - (1^2)] + [(-1/2)(5^2) + 2(5)] - [(-1/2)(1^2) + 2(1)]
Displacement = (250/6 - 25) - (1/6 - 1) + (-25/2 + 10) - (-1/2 + 2)
Displacement = 89/6 + 5/6 + 15 + 5/2
Displacement = 37.83 meters (rounded to two decimal places)
Distance:
To find the distance traveled, we need to integrate the absolute value of the velocity function:
Distance = ∫[1, 5] |v(t)| dt
Distance = ∫[1, 5] [(1/2)t^2 - 2t] dt
Using the power rule for integration, we can find the integral:
Distance = [(1/6)t^3 - t^2] evaluated from 1 to 5
- [(1/6)(1^3) - (1^2)]
Distance = (250/6 - 25) - (1/6 - 1)
Distance = 89/6 + 5/6
Distance = 94/6
Distance = 15.67 meters (rounded to two decimal places)
Therefore, the particle has a displacement of 37.83 meters and travels a distance of 15.67 meters during the time interval 1 ≤ t ≤ 5.