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... :(
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... :(
0 answers