PART A
1. The acceleration of a body along a straight line is given by (a = 2 (ms-3) t -10 (ms-2). Initial position is 1.8 m and initial velocity is 15 ms-1. Determine the position of the particle, the total distance travelled and the velocity after 12 seconds of start.
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substituting t=12s gives
a = 2(12) - 10 = 14m/s-2
Using the formula
d = Vit +1/2 at^2
d = 15(12) + 0.5 (14)12^2
d = 180 + 1008
d = 1188 meters (distance traveled)
add the initial position
d = 1188 + 1.8 = 1189.8 (Position of particle)
I hope this is correct... :(
To determine the position of the particle, total distance traveled, and velocity after 12 seconds, we need to solve them step-by-step using the given information and the given equation for acceleration.
1. Position of the particle (s):
To find the position of the particle at any given time, we need to integrate the equation for acceleration with respect to time (t). Since the equation for acceleration is given as:
a = 2t - 10
Integrating this equation once will give us the equation for velocity:
v = ∫(2t - 10) dt
Applying the integration, we get:
v = t^2 - 10t + C1
Next, we need to find the constant C1 using the initial velocity (v0). Given that the initial velocity is 15 m/s, we can substitute this value into the velocity equation:
15 = (0)^2 - 10(0) + C1
C1 = 15
Now, we have the equation for velocity:
v = t^2 - 10t + 15
To find the position (s), we need to integrate the equation for velocity with respect to time:
s = ∫(t^2 - 10t + 15) dt
Applying the integration, we get:
s = (1/3)t^3 - 5t^2 + 15t + C2
Again, we need to find the constant C2 using the initial position (s0). Given that the initial position is 1.8 m, we can substitute this value into the position equation:
1.8 = (1/3)(0)^3 - 5(0)^2 + 15(0) + C2
C2 = 1.8
Now, we have the final equation for the position:
s = (1/3)t^3 - 5t^2 + 15t + 1.8
2. Total distance traveled:
To find the total distance traveled, we need to consider both the positive and negative displacements. When the particle is moving in the positive direction, the displacement contributes to the total distance traveled. When the particle is moving in the negative direction, the displacement subtracts from the total distance traveled.
To find the total distance traveled, we first determine the time at which the particle changes direction. This happens when the velocity (v) equals zero:
t^2 - 10t + 15 = 0
Solving this quadratic equation, we find:
t = 5 or t = 3
Since the initial time is 0, we have two time intervals: t = [0, 3] and t = [3, 12].
For the time interval t = [0, 3]:
We substitute the limits of integration into the position equation:
s1 = (1/3)(3)^3 - 5(3)^2 + 15(3) + 1.8
s1 = 9 - 45 + 45 + 1.8
s1 = 10.8 m
For the time interval t = [3, 12]:
We substitute the limits of integration into the position equation:
s2 = (1/3)(12)^3 - 5(12)^2 + 15(12) + 1.8
s2 = 576 - 720 + 180 + 1.8
s2 = 37.8 m
The total distance traveled is the sum of the absolute values of the displacements:
Total distance = abs(s1) + abs(s2)
Total distance = abs(10.8) + abs(37.8)
Total distance = 10.8 + 37.8
Total distance = 48.6 m
3. Velocity after 12 seconds:
To find the velocity after 12 seconds, we substitute the value of t = 12 into the equation for velocity:
v = (12)^2 - 10(12) + 15
v = 144 - 120 + 15
v = 39 m/s
Therefore, the position of the particle after 12 seconds is 37.8 m, the total distance traveled is 48.6 m, and the velocity is 39 m/s.