# The position of a particle along the x axis depends on the time according to the equation x = At2 - Bt3, where x is in meters and tis in seconds. (a) What SI units must A and B have? For the following, let their numerical values in SI units be 3.0 and 1.0, respectively. (b) At what time does the particle reach its maxi- mum positive x position? (c) What total path-length does the particle cover in the first 4 seconds? (d) What is its displacement during the first 4 seconds? (e) What is the particle's velocity at the end of each of the first 4 seconds? (j) What is the particle's acceleration at the end of each of the first 4 seconds? (g) What is the average velocity for the time interval t = 2 to t = 4 s?

Navjyot Kalra
9 years ago
Given:
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

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:

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:

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.

Thus the path length of the particle in first 4 seconds is 26 m.
(d) The path length in the part above accounts for the total distance travelled by the particle in first 4 seconds. However to account for its displacement, one has to calculate the particle position relative to its initial point.
As said above, the particle reaches its initial position at t = 3 s, therefore the net displacement of the particle is calculated from the particle position at t = 4 s which is -16 m.
Therefore the net displacement of the particle in first four seconds is - 16m.
(e) The velocity (say v) of the particle at any instant is given by differentiation of its position x with respect to time as:

a = 2A – 6Bt ….….….….….….2

deddhy
19 Points
7 years ago
represent the distance, x meters, moved by a body in t seconds. Determine :
1. The velocity and acceleration at the start
2. The velocity and accelaration  when t = 3 s
3. The values of  t when the body is at rest
4. The value of   t when the accelation is 37 m/s2
5. The distance travelled in  the third second