The position of a particle moving along thex-axis depends on the time according to the equation x = at2
– bt3, where x is in feet and t in second. For the following, let the numerical values of a and b be 3.0
and 1.0, respectively. (a) At what time does the particle reach its maximum positive x-position ?
(b) What total length of path does the particle cover in the first 4.0 s ? (c) What is its displacement dur-
ing the first
4.0 s ?
(d) What is the particle’s speed at the end of each of the first four seconds ?
(e) What is the particle’s acceleration at the end of each of the first four seconds
Answers
Answer:
Position of particle, x = At2 – Bt2.
Numerical value of coefficient , A, A = 3.0 m/s2
Numerical value of coefficient , B , B = 1.0/s3
For the equation given as x = At2 – Bt3 to be dimensionally correct, the dimensions on the left side must be equal to the right. However this is true if the dimensions of At2 and Bt3 matches that of x.
Since x is measured in meters in S.I units, therefore the dimensions of At2 and Bt3 should also be meters.
The dimensions of t in S.I units is seconds, therefore the dimension of A must be meters/ second2. Thus one can see that the multiplication of A and t2 will bring a quantity with dimensions of .
Therefore the dimensions of A is meters/second2.
Similarly, if the dimension of B is meters/second3, then on multiplication with t3 it will produce a quantity with dimensions of .
Therefore the dimension of B is meters/second3
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Therefore we find that the last time the object was at positive x axis was just before 3s.
One can substitute various value of time t to see the behavior of the object. It is clear from the equation that the particle starts atx = 0 m, moves towards the positive x axis with time. The maximum value of x occurs at t = 2 sand is given as:
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The particle has moved 4 m in the first 2 seconds, then reversed its direction and moved along the negative x axis. At t = 3 s, the particle managed to reached the position from where it started its motion.Therefore the total distance travelled by the particle till 3 seconds is twice the maximum displacement towards positive x axis that is 2(4 m) = 8 m.
The position of the particle at t = 4 s is given as:
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The particle has went to x = 16 m towards the negative x axis from t = 3 s to t = 4s. Therefore the total distance travelled by the particle till 4 seconds is the sum of the distance travelled in first 3 seconds, which is and the distance travelled by it in the next second, which is 16m.
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Answer:
X = at^2 + bt^3If x is in meters then at^2 and bt^3 should also be ... time according to equation x=at2 + bt3 where,x is in meter and t in seconds.What are the unit and dimension of a and b? 2.