Calculate distance traveled with opposite force being applied to item travelling

Question

An item moves for $20$ seconds on a straight path with a beginning velocity of $20m/s$. For this entire duration, a force $(X)$ is added to the item which results in this item decelerating at $6m/s^2$. What is the items’ distance from its initial point after the entire 20 seconds have passed?

I think, distance = (speed)(time) = (20)(20) = 400m.

And also, distance = (speed)(time) = (6)(20) = 120m.

Therefore, distance traveled over 20 seconds is 400m – 120m = 280m.

Is this correct? I am worried that I might be oversimplifying this.

Thanks.

Answer 1

Let apply

$$s=v_ot+\frac12at^2$$

with

• $v_0=20 m/s$
• $a=-6 m/s^2$
• $t=20 s$

$$\implies s=v_ot+\frac12at^2=400-\frac126\cdot400=-800 \,m$$

Answer 2

You have a particle moving with constant acceleration so you need to use suvat equations

In this case you have $a=-6,u=20,t=20$ and you can use the equation $s=ut+\frac 12at^2$ to get $s=-800$ so the distance from the initial point is $800$

Answer 3

With physicist notations: $$F=m\ddot{r}$$ $$\frac{F}{m}=\ddot{r}$$ $$a=\ddot{r}$$ $$-6=\ddot{r}$$ $$-6=\dot{v}$$ Integrating both sides with respect to $t$: $$\int\limits_{t=0}^{t}-6\mathrm{d}t=\int\limits_{v=20}^{v}\mathrm{d}v$$ $$-6t=v-20$$ $$v(t)=20-6t$$ $$\dot{r}=v$$ $$r(t)=\int\limits_{t=0}^{t}20-6t\mathrm{d}t$$ $$r(t)=20t-3t^2$$ You are looking for $r(20)$: $$r(20)=20*20-3*20^2=-800\text{m}$$ So it will be $800$m away from the initial position.