Physics related antiderivatives problem

Question

I was wondering if I could get a little help with a calculus related word problem:

A car braked with a constant deceleration of 5 meters per second squared for 60 meters before stopping. How fast was the car traveling when the brakes were applied? I can't use the definite integral, only the antiderivative.

Thanks for any help!

Answer 1

By definition the acceleration is $a = \dot v$, where $v$ is the velocity and $\dot{\phantom{v}}$ denotes derivative with respect to time. If $a$ is constant we can take the antiderivative to get $$v(t) = at + v_0$$ where $v_0$ is the velocity at time $t=0$.

Now $v = \dot s$ where $s$ is the distance travelled. Using the above expression for $v$ we can again take the antiderivative and get $$s(t) = \frac12at^2 + v_0t + s_0$$ where $s_0$ is the distance travelled at $t=0$.

Assume that the driver applies the brakes at time $t_0$ and that the car stops at time $t_1$. Then we have $$\Delta s = s(t_1) - s(t_0) = (\frac12at_1^2 + v_0t_1 + s_0) - (\frac12at_0^2 + v_0t_0 + s_0) = \frac12a(t_1^2-t_0^2) + v_0(t_1-t_0) \tag{1}$$ where $\Delta s$ is the distance travelled during braking, and also $$0 = v(t_1) = at_1 + v_0 \tag{2}$$

In the exercise $\Delta s = 60\,\text{m}$ and $a = -5 \text{m/s}^2$ are given and we are asked for $v(t_0)$.

Using (2) we get $t_1 = -v_0/a$, which inserted into (1) gives $$\Delta s = \frac12a((-v_0/a)^2-t_0^2) + v_0(-v_0/a-t_0) = -\frac12v_0^2/a - v_0t_0 - \frac12at_0^2 = -\frac12(v_0+at_0)^2/a = -\frac12v(t_0)^2/a$$

Thus, $$v(t_0) = \pm\sqrt{2a\Delta s}$$ Now you can just insert the given values of $a$ and $\Delta s$.

I have in these calculations deliberately not taken $t_0=0$.

Answer 2

You just need the formula

$$V_1^2-V_2^2=2\gamma (X_1-X_2) $$

with $V_i =$ speed at position $X_i $ and $\gamma $ the constant acceleration.

thus

$$V_1^2-0=2.5.60$$

$$V_1=\sqrt {600} m/s$$

The formula comes from the integration of

$$\frac {d (V^2/2)}{dt}=V\frac {dV}{dt}=\gamma V=\gamma \frac {dX}{dt} $$

Answer 3

Let $travel_m = as^2+bs+c$ (s = time, travel_m = travelled distance)

$\frac{\text{d}^2y}{\text{d}s^2} as^2+bs+c$ (The second derivative of distance gives you the acceleration.)

$=2a=-5$

$a=-2.5$

$travel_m = -2.5s^2+bs+c$

At first, the travel distance was $0$, $c=0$.

$travel_m = -2.5s^2+bs$

$\frac{\text{d}y}{\text{d}s}-2.5s^2+bs$ (The first derivative of distance gives you the speed.)

$=-5s + b$, but at $s=0$, when the car braked, the speed was b, and they are asking for b.

When it stopped the speed was zero. To see when it stopped:

$-5s + b=0$ -> $\frac{b}{5}=s$

$-2.5s^2+bs = 60$

Substitute: $-\frac{5}{2}\times\frac{b^2}{25}+\frac{b^2}{5} = 60$ -> $b^2=600$

$\boxed{b=10\sqrt6}$