# An object is moving with initial velocity of 20m/s. It must slow down to a stop in .33 seconds and travel 4m. Write a solution that applies a force to the object to satisfy the expected result.?

## The first thing I do not understand is why I can assume linear acceleration. I have a basic understanding of calculus so explaining using integration is fine.

Jan 4, 2017

$F = m \cdot 36.7 \frac{m}{s} ^ 2$

#### Explanation:

You may “assume” linear acceleration because you are asked for “a solution”, not all solutions.

F = m * a
d = v * t To stop in the specified distance and time.
4m = v(m/s) * 0.33s ; v = 12.12 m/s

To decelerate:
a = (v/t) ; a = ((12.12(m/s))/0.33s) = 36.7 m/s^2

F = m * a

$F = m \cdot 36.7 \frac{m}{s} ^ 2$

Jan 6, 2017

Applied force $F \left(t\right) = \left(- 22.0 m - 233.7 m t\right)$ N

#### Explanation:

Initial velocity $u = 20 m {s}^{-} 1$. Final velocity $v = 0$, distance traveled $= 4 m$.

The applicable kinematic equation for a constant acceleration is
${v}^{2} - {u}^{2} = 2 a s$
$\implies {0}^{2} - {20}^{2} = 2 a \times 4$
$\implies a = - \frac{400}{8} = - 50 m {s}^{-} 2$

Let us find out for what time this retarding force is applied.
Using the other kinematic equation
$v = u + a t$
we get
$0 = 20 + \left(- 50\right) t$
$t = \frac{20}{50} = 0.4 s$
We see that the object takes more time than given in the question to come to stop. This implies that it is not a case of constant deceleration.

Let us assume that the object of mass $m$ is acted upon by a time dependent force $F \left(t\right)$. From Newton's Second law we get
$F \left(t\right) = m \times a \left(t\right)$ .....(1)
We know that acceleration can be written as
$\frac{\mathrm{dv}}{\mathrm{dt}} = a \left(t\right)$
$\implies \mathrm{dv} = a \left(t\right) \cdot \mathrm{dt}$
Using (1) and Integrating both sides we get
$v \left(t\right) = \int \frac{F \left(t\right)}{m} \cdot \mathrm{dt}$......(2)

Suppose the force is given by the expression
$F \left(t\right) = L + K t$
Where $L \mathmr{and} K$ are constants. (2) becomes
$v \left(t\right) = \int \frac{L + K t}{m} \cdot \mathrm{dt}$
$\implies v \left(t\right) = \frac{L}{m} t + \frac{K}{m} {t}^{2} / 2 + C$
where $C$ is constant of integration.
As $v \left(t\right) = u$ initial velocity at $t = 0$
Expression becomes

$v \left(t\right) = \frac{L}{m} t + \frac{K}{2 m} {t}^{2} + u$ .....(3)
We also know that velocity can be written in terms of displacement $s$ as
$\frac{\mathrm{ds}}{\mathrm{dt}} = v \left(t\right)$
$\implies \mathrm{ds} = v \left(t\right) \cdot \mathrm{dt}$
Integrating both sides we get
$s = \int v \left(t\right) \cdot \mathrm{dt}$
Using (3) we get
$s = \int \left(\frac{L}{m} t + \frac{K}{2 m} {t}^{2} + u\right) \cdot \mathrm{dt}$
$\implies s = \frac{L}{m} {t}^{2} / 2 + \frac{K}{2 m} {t}^{3} / 3 + u t + {C}_{1}$
${C}_{1}$ is constant of integration. Assuming that displacement is zero at $t = 0$, we get ${C}_{1} = 0$ and the expression for displacement becomes
$\implies s = \frac{L}{2 m} {t}^{2} + \frac{K}{6 m} {t}^{3} + u t$ ......(4)

Inserting given values in (3) and (4) we obtain
$0 = \frac{L}{m} \times 0.33 + \frac{K}{2 m} {\left(0.33\right)}^{2} + 20$ .....(5)
$4 = \frac{L}{2 m} {\left(0.33\right)}^{2} + \frac{K}{6 m} {\left(0.33\right)}^{3} + 20 \times 0.33$ ......(6)
To solve for $L \mathmr{and} K$, multiply (5) with $\frac{0.33}{2}$, it becomes
$0 = \frac{L}{2 m} \times {\left(0.33\right)}^{2} + \frac{K}{4 m} {\left(0.33\right)}^{3} + 20 \times \frac{0.33}{2}$ .....(7)

Subtracting (7) from (6)
$4 = \frac{K}{6 m} {\left(0.33\right)}^{3} + 20 \times 0.33 - \left(\frac{K}{4 m} {\left(0.33\right)}^{3} + 20 \times \frac{0.33}{2}\right)$
$\implies 4 = - \frac{K}{12 m} {\left(0.33\right)}^{3} - 20 \times \frac{0.33}{2}$
$K = - 233.7 m$
Inserting this value of $K$ in (5) we get
$0 = \frac{L}{m} \times 0.33 - \frac{233.7 m}{2 m} {\left(0.33\right)}^{2} + 20$
$L = - 22.0 m$

As such applied force $F \left(t\right) = \left(- 22.0 m - 233.7 m t\right)$ N