A particle moving along a straight line with a constant acceleration of -4m/s*2 passes through a point A with a velocity of +8m/s at some movement find the distance travelled by the particle in 5sec after that movement?

Ans:26m

To find the distance travelled by the particle in 5 seconds after a certain point, we can use the equations of motion for uniformly accelerated motion.

The first equation of motion is:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time

In this case, we know that the initial velocity (u) is +8 m/s, the acceleration (a) is -4 m/s^2, and the time (t) is 5 seconds. Therefore, we can calculate the final velocity (v) using the equation above.

v = u + at
v = 8 + (-4) * 5
v = 8 - 20
v = -12 m/s

Now that we know the final velocity after 5 seconds, we can use the second equation of motion to find the displacement (distance travelled) during this time interval.

The second equation of motion is:
s = ut + (1/2)at^2
where:
s = displacement (distance travelled)
u = initial velocity
t = time
a = acceleration

Using the given values, we can calculate the displacement:

s = ut + (1/2)at^2
s = 8 * 5 + (1/2) * (-4) * (5^2)
s = 40 + (-10)
s = 30 m

Therefore, the particle travels a distance of 30 meters in 5 seconds after the given movement.