The equation of motion of the body has the form s = 20t-0.4t ^ 2 1) determine how many seconds from

The equation of motion of the body has the form s = 20t-0.4t ^ 2 1) determine how many seconds from the starting point the body will stop? 2) the module of acceleration “a” with which the body moves?

Given:

s (t) = 20 * t – 0.4 * t ^ 2 – the equation of motion of a body.

It is required to determine t (second) – the time interval after which the body will stop, as well as a (m / s2) – the body’s acceleration.

Let us find the equation of the body’s velocity by fulfilling the first-degree derivative of the equation of motion:

v (t) = s (t) ’= (20 * t – 0.4 * t ^ 2)’ = 20 – 0.8 * t.

When the body stops, its speed will be zero. Based on this, we get:

20 – 0.8 * t = 0;

20 = 0.8 * t;

t = 20 / 0.8 = 200/8 = 25 seconds.

Let us find the acceleration of the body by fulfilling the derivative of the velocity equation:

a = v (t) ’= (20 – 0.8 * t) = -0.8 m / s2, | a | = | -0.8 | = 0.8 m / s2.

Answer: the body will stop 25 seconds after the start of movement, the body’s acceleration modulus is 0.8 m / s2.