Question

A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. Calculate how far the car travels in the time it takes to stop. Round your answer to one decimal place.

Answer 1

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ($v_0$) = 40 ft/sec

Deceleration of the car ($\frac{dv}{dt}$) = -10 ft/sec²

Final speed of the car ($v_x$) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.