# force and motion

+1 vote
A particle moves along a straight line such that its displacement during  any given interval of 1 sec is 3 metres larger than its displacement during the previous interval of 1 second.which of the following is correct-

1.the particle moves with constant acceleration of 3m/s^2

2.the particle moves with constant velocity of 3m/s

3.the particle moves with constant velocity of 6m/s.

4.the acceleration of the particle is increasing with time.

which is correct?
asked Oct 28, 2013 in Science

+1 vote

If you think about it: let's say the given is X=0 at t=0. We can use the table below.

time(t) distance (x)
0s 0
1s 3
2s 9
3s 18

Now, we can see that the distance increases at an exponential rate, so these kinds of data would not be a result of a particle moving at a constant velocity. If the velocity is constant at 3m/s, our distance would be 0,3,6,9. We can use this kind of thinking for choice 3. If the velocity is constant at 6m/s, then the distance would 0,6,12,18 in 4 seconds. We have now eliminated choices 2 and 3.

To eliminate between choices 1 and 4, we can use one of our kinematic equations (x=vi*t+0.5*a*t^2). At t = 3, our x should be 18 if our initial velocity is equal to 0.

x=0+0.5*(3m/s^2)(3s)^2

x=27/2

x is not equal to 18 which means choice 1 is false. This makes choice 4 the only correct answer. Also, we know our answer is correct because the velocity increases exponentially as time t increases.

answered Nov 3, 2013 by ~Expert~ (1,508 points)

## Related questions

1 answer 29 views
29 views