Question

Motion of a particle is given by equation s= (3 t cube+ 7 t square +14 t + 8 )m the value of acceleration of the particle at t=1s is ???????????????????????????

Answer 1

Motion of a particle is given by equation $s = (3t \text{ cube} + 7t \text{ square} + 14t + 8)m$ the value of acceleration of the particle at $t = 1s$ is?

**Solution.**

$$s = (3t^3 + 7t^2 + 14t + 8)m, t = 1s;$$ $$a - ?$$ $$s = 3t^3 + 7t^2 + 14t + 8.$$

The velocity is the derivative of the displacement vector as a function of time:

$$v = \frac{ds}{dt} = 9t^2 + 14t + 14.$$ $$v = (9t^2 + 14t + 14)\frac{m}{s}.$$

The acceleration is the derivative of the velocity vector as a function of time:

$$a = \frac{dv}{dt} = 18t + 14.$$ $$a = (18t + 14)\frac{m}{s^2}.$$

The value of acceleration of the particle at $t = 1s$:

$$a = 18 \cdot 1 + 14 = 32.$$ $$a = 32\frac{m}{s^2}.$$

**Answer:** The value of acceleration of the particle at $t = 1s$ is $a = 32\frac{m}{s^2}$.