A force acts on a 2 kg particle whose position is given by x = (2t-t^2+3) m. The work done on the particle in first 4s is options
Answers
Answered by
0
Explanation:
We're asked to find the distance traveled by the object after the first three seconds of its motion, given its one-dimensional position equation.
Notice that it doesn't ask for the displacement of the particle, it asks for the total distance traveled. With that in mind, we can't simply plug in the t=0 and t=3 s values and have that be the answer, we would have to substitute allvalues for t:
Since it's asking for the distance traveled, we can disregard the "+6" quantity in the equation, because at time t=0 it obviously hasn't gone anywhere yet. The process a as follows:
x(0)=(0)2−4(0)=0 m
x(0)=(1)2−4(1)=−3 m
x(2)=(2)2−4(2)=−4 m
x(3)=(3)2−4(3)=−3 m
We can see that the particle traveled 4 meters in the negative direction, and then one meter in the positive direction. The total distance it traveled was
4 m +1 m =5 m
so option (1) is correct.
We're asked to find the distance traveled by the object after the first three seconds of its motion, given its one-dimensional position equation.
Notice that it doesn't ask for the displacement of the particle, it asks for the total distance traveled. With that in mind, we can't simply plug in the t=0 and t=3 s values and have that be the answer, we would have to substitute allvalues for t:
Since it's asking for the distance traveled, we can disregard the "+6" quantity in the equation, because at time t=0 it obviously hasn't gone anywhere yet. The process a as follows:
x(0)=(0)2−4(0)=0 m
x(0)=(1)2−4(1)=−3 m
x(2)=(2)2−4(2)=−4 m
x(3)=(3)2−4(3)=−3 m
We can see that the particle traveled 4 meters in the negative direction, and then one meter in the positive direction. The total distance it traveled was
4 m +1 m =5 m
so option (1) is correct.
Similar questions