a particle moves with a constant acceleration of 2 m/s^2 in x direction. The distance travelled in fifth second is 19 m. calculate the distance travelled after 5 second.

Let u be the particle's initial velocity. Then the distance traveled in the first 5 seconds is such that

19 m = u (5 s) + 1/2 (2 m/s²) (5 s)²

Solve for u :

u = (19 m - 1/2 (2 m/s²) (5 s)²) / (5 s) = - 1.2 m/s

That is, the particle was moving in the negative x direction with speed 1.2 m/s. The particle's velocity after 5 seconds is then

v = u + (2 m/s²) (5 s) = 8.8 m/s

So, in the following 5-second interval, assuming it maintains the same acceleration, the particle travels a distance of

x = v (5 s) + 1/2 (2 m/s²) (5 s)² = 69 m

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