# The body moves in a straight line according to the law S (t) = 21t to the power 3+ 21t ^ 2 + 7t-2

**The body moves in a straight line according to the law S (t) = 21t to the power 3+ 21t ^ 2 + 7t-2 find the acceleration speed at the end of 10 seconds V (t), a (t), v (10), a (10)**

Let’s calculate the equation of speed change – it is equal to the time derivative of the path:

v (t) = s’ (t) = 63 * t² + 42 * t + 7.

Speed 10 seconds after starting movement:

v (10) = 63 * 10² + 42 * 10 + 7 = 6727 units / s.

Let’s find the law of acceleration change – it is equal to the derivative of the speed with respect to time:

a (t) = v ‘(t) = 126 * t + 42.

Acceleration in 10 seconds from the moment of the start of movement:

a (10) = 126 * 10 + 42 = 1302 units / s².

The rate of change of acceleration is the derivative of the acceleration with respect to time, i.e .:

va (t) = a ‘(t) = 126, i.e. the value is constant, i.e. acceleration changes at the same speed.