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The position of a particle along the x axis depends on the time according to the equation x = At^{2} - Bt^{3}, 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?

6 years ago

Position of particle, x = At

Numerical value of coefficient , A, A = 3.0 m/s

Numerical value of coefficient , B , B = 1.0/s

For the equation given as x = At

Since x is measured in meters in S.I units, therefore the dimensions of At

The dimensions of t in S.I units is seconds, therefore the dimension of A must be meters/ second

Therefore the dimensions of A is meters/second

Similarly, if the dimension of B is meters/second

Therefore the dimension of B is meters/second

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 at

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

5 years ago

represent the distance, x meters, moved by a body in t seconds. Determine :

- The velocity and acceleration at the start
- The velocity and accelaration when t = 3 s
- The values of t when the body is at rest
- The value of t when the accelation is 37 m/s
^{2 } - The distance travelled in the third second

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