a body is moving along +xaxis with uniform acceleration of-4ms2 its velocity at x=0 is 10ms-1 the time taken by the body to reach a point at x=12m is?

To find the time taken by the body to reach a point at x = 12m, we can use the equation of motion:

x = ut + (1/2)at^2

Where:
x is the displacement (12m in this case).
u is the initial velocity (10 m/s in this case).
a is the acceleration (-4 m/s^2 in this case).
t is the time taken.

Rearranging the equation, we have:

12 = 10t + (1/2)(-4)t^2

Multiplying all terms by 2 to eliminate fractions:

24 = 20t - 4t^2

Rearranging the equation and setting it to zero:

4t^2 - 20t + 24 = 0

Dividing the equation by 4:

t^2 - 5t + 6 = 0

Now, we can factorize the quadratic equation:

(t - 3)(t - 2) = 0

Setting each factor to zero, we have two possible solutions for t:

t - 3 = 0 ---> t = 3s
t - 2 = 0 ---> t = 2s

However, since we are looking for the time taken by the body to reach a point at x = 12m, the time cannot be negative. Therefore, the correct solution is:

t = 3s

So, the body takes 3 seconds to reach the point at x = 12m.

To find the time taken by the body to reach a point at x = 12m, we can use the kinematic equation:

x = x0 + (v0 * t) + (0.5 * a * t^2)

Given:
Initial position (x0) = 0
Initial velocity (v0) = 10 m/s
Acceleration (a) = -4 m/s^2
Final position (x) = 12 m

We need to find time (t).

Let's substitute the given values into the equation:

12 = 0 + (10 * t) + (0.5 * (-4) * t^2)

Simplifying the equation:

12 = 10t - 2t^2

Rearranging the equation to make it quadratic:

2t^2 - 10t + 12 = 0

Now, we can solve this quadratic equation to find the values of 't'.

Using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

For the equation 2t^2 - 10t + 12 = 0, the values of a, b, and c are:

a = 2, b = -10, c = 12

Substituting these values into the quadratic formula:

t = (-(-10) ± sqrt((-10)^2 - 4 * 2 * 12)) / (2 * 2)

Simplifying:

t = (10 ± sqrt(100 - 96)) / 4
t = (10 ± sqrt(4)) / 4
t = (10 ± 2) / 4

There are two possible solutions:

t1 = (10 + 2) / 4 = 12 / 4 = 3 seconds
t2 = (10 - 2) / 4 = 8 / 4 = 2 seconds

Therefore, the time taken by the body to reach a point at x = 12m is either 3 seconds (t1) or 2 seconds (t2).

d = Vo*t + 0.5*t^2.

12 = 10*t + 2t^2,
2t^2 + 10t - 12 = 0.
t^2 + 5t - 6 = 0.
(t-1)(t+6) = 0,
t-1 = 0, t = 1 s.
t+6 = 0, t = -6.
Solution: t = 1 s.