Q. 41 A particle鈥檚 acceleration is d... [FREE SOLUTION]

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Chapter 2: Q. 41 (page 61)

A particle鈥檚 acceleration is described by the function ax = 110 - t2 m/s2 , where t is in s. Its initial conditions are x0 = 0 m and v0x = 0 m/s at t = 0 s. a. At what time is the velocity again zero? b. What is the particle鈥檚 position at that time?

Short Answer

Expert verified

Part (a)

The particle will have zero velocity again at $t = 20 s$ .

Part (b)

The position of the particle at $t = 20 s$ is $667 m$ .

Step by step solution

Given information

An acceleration of the particle is given by the function $a_{x} = (10 - t) \frac{m}{s^{2}}$ .

Initial conditions are $x_{0} = 0 m$ , $v_{0 x} = 0 m / s$ ,and $t = 0 s$ .

Part (a)

The velocity is given by

$v_{f} = v_{0 x} + {鈭�}_{t_{0}}^{t_{f}} a_{x} d t = v_{0 x} + {鈭�}_{t_{0}}^{t_{f}} (10 - t) d t = v_{0 x} + {[10 t - \frac{t^{2}}{2}]}_{t_{0}}^{t_{f}}$

Substitute the known values

$v_{1} = v_{0 x} + {[10 t - \frac{t^{2}}{2}]}_{0}^{t_{1}} 0 = 0 + (10 t_{1} - \frac{t_{1}^{2}}{2}) t_{1}^{2} - 20 t_{1} = 0 t_{1} = 0 s o r t_{1} = 20 s$

Therefore, the particle will have zero velocity at $t_{1} = 20 s$ .

Part (b)

The position of the particle is

$x_{f} = x_{0} + {鈭�}_{t_{0}}^{t_{f}} v_{x} d t = x_{0} + {鈭�}_{t_{0}}^{t_{f}} (10 t - \frac{1}{2} t^{2}) d t = x_{0} + {[5 t^{2} - \frac{t^{3}}{6}]}_{t_{0}}^{t_{f}}$

Substitute the known values

$x_{1} = x_{0} + {[5 t^{2} - \frac{t^{3}}{6}]}_{t_{0}}^{t_{1}} x_{1} = x_{0} + {[5 t^{2} - \frac{t^{3}}{6}]}_{0}^{20} = 0 + 5 {(20)}^{2} - \frac{20^{3}}{6} = 667 m$