Question

a car is rolling back when it hits the gas. after 8.25s, it is moving forward at 8.62m/s, and is 12.9m to the right of the starting point. what was its starting velocity?

Answer 1

Answer on Question #46137, Physics, Mechanics | Kinematics | Dynamics

A car is rolling back when it hits the gas. After 8.25 s, it is moving forward at 8.62 m/s, and is 12.9 m to the right of the starting point. What was its starting velocity?

Solution:

The kinematic equation that describes an object's motion is:

$$x = x _ {0} - v _ {0} t + \frac {1}{2} a t ^ {2}$$

where

$x_0 = 0$ is initial position

$v_{0} = ?$ is initial speed

$a$ is acceleration

At time $t = 8.25$ s the position of car is $x = 12.9$ m.

Thus, from first equation

$$v _ {0} = \frac {a t}{2} - \frac {x}{t}$$

The acceleration is

$$a = \frac {v - v _ {0}}{t}$$

In our case, the initial velocity has minus sign.

Thus,

$$a = \frac {v - (- v _ {0})}{t} = \frac {v + v _ {0}}{t}$$

Substituting

$$v _ {0} = \frac {t}{2} \left(\frac {v}{t} + \frac {v _ {0}}{t}\right) - \frac {x}{t} = \frac {v}{2} + \frac {v _ {0}}{2} - \frac {x}{t}$$

Thus,

$$\frac {v _ {0}}{2} = \frac {v}{2} - \frac {x}{t}$$

So,

$$v _ {0} = v - \frac {x}{t} = 8.62 - \frac {12.9}{8.25} = 7.06 \mathrm{~m/s}$$

Answer: $v_{0} = 7.06 \, \mathrm{m/s}$

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