Calculation of the velocity of an object

Question

I have the position function $$s(t) = -4.9t^2 + v_0t + s_0$$ for free falling objects. The question is what is the velocity of an object after $5$ seconds with initial velocity $120$ m/s.

I tried to find the position $s(5)$ of the object after it will pass $5$ seconds $$s(5) = -4.9\cdot 25 + 120 \cdot 5 + 0 = 447.5$$ and from here I get the velocity $s(5)/5$ which is $95.5$ (in my book is $71$ only) then I am asked to do the same thing for $10$ seconds here I get the result $59$ and in my book it is $22$ m/s

Answer 1

Hint: The velocity at a certain $t_0$ is given by $v(t_0) = \frac{ds}{dt}(t_0)$. Where $$\frac{ds}{dt} (t_0) = \lim_{h \to 0}\frac{s(t_0 + h) - s(t_0)}{h}$$

Now what is $v(5)$?

Answer 2

The equation for change in velocity by time is $$\Delta v(t) = a * t\\v_2-v_1=at$$ Here we have $a=-4.9 \frac{m}{s^2}$, we have $t=5s$ and we have $v_1=120 \frac{m}{s}$. This means that our final time $v_2$ will be $$v_2 = -4.9\frac{m}{s^2}(5s) + 120\frac{m}{s}\\v_2 = -24.5\frac{m}{s}+120\frac{m}{s}\\v_2 = 95.5\frac{m}{s}$$ At $t=10s$ the answer is $$v_2 = -4.9\frac{m}{s^2}(10s) + 120\frac{m}{s}\\v_2 = -49\frac{m}{s}+120\frac{m}{s}\\v_2 = 71\frac{m}{s}$$

Answer 3

Here are two approaches.

Method 1: Using calculus.

The vertical position of the object at time $t$ is given by the formula $$s(t) = \frac{1}{2}gt^2 + v_0t + s_0$$ where $g$ is the acceleration due to gravity in the Earth's gravitational field, $v_0$ is the initial velocity, and $s_0$ is the initial position.

The instantaneous velocity is the derivative of the position function. Hence, the velocity at time $t$ is $$v(t) = gt + v_0$$ To find the velocity at $t = 5~\text{s}$, we substitute $-9.8~\text{m}/\text{s}^2$ for $g$, $5~\text{s}$ for $t$, and $120~\text{m}/\text{s}$ for $v_0$, which yields $$v(5) = \left(-9.8~\frac{\text{m}}{\text{s}^2}\right)(5~\text{s}) + 120~\frac{\text{m}}{\text{s}} = -49~\frac{\text{m}}{\text{s}} + 120~\frac{\text{m}}{\text{s}} = 71~\frac{\text{m}}{\text{s}}$$ To find the velocity at $t = 10~\text{s}$, substitute $10~\text{s}$ for $t$ instead.

Method 2: Without calculus.

Since the acceleration is constant, the vertical component of the velocity at time $t$ is given by the formula $$v(t) = v_0 + gt$$ where $v_0$ is the initial velocity and $g$ is the acceleration due to gravity in the Earth's gravitational field. Note that this is the formula we obtained above by differentiating the position function with respect to time.

To find the velocity at $t = 5~\text{s}$, substitute $120~\text{m}/\text{s}$ for $v_0$, $-9.8~\text{m}/\text{s}^2$ for $g$, and $5~\text{s}$ for $t$, which yields $$v(5) = 120~\frac{\text{m}}{\text{s}} + \left(-9.8~\frac{\text{m}}{\text{s}^2}\right)(5~\text{s}) = 120~\frac{\text{m}}{\text{s}} - 49~\frac{\text{m}}{\text{s}} = 71~\frac{\text{m}}{\text{s}}$$ To find the velocity at $t = 10~\text{s}$, substitute $10~\text{s}$ for $t$ instead.