the velocity of the body moving in a straight line with a constant acceleration, is 10 m/s at a certain instant 't' . after 5 second the velocity become 20 m/s, the velocity 3 second before 't' was?

To find the velocity 3 seconds before the certain instant 't', we can use the equations of motion.

Let's assume the velocity 'v0' at time 't0' seconds is the initial velocity, and 'v' at time 't' seconds is the final velocity. We are given that:
v0 = 10 m/s (at time t0)
v = 20 m/s (after 5 seconds, at time t)

Using the equation of motion for constant acceleration:
v = v0 + at

Rearranging the equation to solve for 'a', the acceleration:
a = (v - v0) / t

Substituting the given values:
a = (20 m/s - 10 m/s) / 5 s
a = 10 m/s / 5 s
a = 2 m/s²

Now, we can find the velocity 3 seconds before time 't'. Let's call it v3.

Using the equation of motion again:
v = v0 + at

Substituting the known values:
20 m/s = v0 + 2 m/s² * 5 s

Rearranging the equation to solve for v0, the initial velocity:
v0 = 20 m/s - 2 m/s² * 5 s
v0 = 20 m/s - 10 m/s
v0 = 10 m/s

Therefore, the velocity 3 seconds before time 't' was 10 m/s.

To find the velocity 3 seconds before the certain instant 't', we need to use the equations of motion for an object moving in a straight line with constant acceleration.

The equation that relates velocity (v) to time (t) and acceleration (a) is:
v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given information:
At a certain instant 't': v = 10 m/s
After 5 seconds: v = 20 m/s

Now, let's find the acceleration (a) using the given information. We can rearrange the equation to solve for acceleration:
a = (v - u) / t

Using the first set of given information:
10 = u + at

Using the second set of given information:
20 = u + a(5)

Now, we can solve these two equations simultaneously to find the values of 'u' and 'a'. Subtracting the first equation from the second, we get:
(20 - 10) = u + a(5) - (u + at)
10 = 5a

Thus, we have:
a = 2 m/s²

Substituting the value of 'a' into the first equation:
10 = u + 2t

Substituting t = 5 into the equation, we can solve for 'u':
10 = u + 2(5)
10 = u + 10
u = 0 m/s

So, the initial velocity (3 seconds before 't') is 0 m/s.

clearly, the acceleration is ((20-10)m/s)/((5-0)s) = 10/5 = 2 m/s^2

so at t-3, v = 10-3*2 = 4 m/s