Question

Calculate the acceleration of a 1400-kg car that stops from 39 km/h "on a dime" (on a distance of 1.7 cm).

Answer 1

Answer:

$a=-3449.67\frac{m}{s^2}$

Explanation:

The car is under an uniforly accelerated motion. So, we use the kinematic equations. We calculate the acceleration from the following equation:

$v_f^2=v_0^2+2ax$

We convert the initial speed to $\frac{m}{s}$

$39\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=10.83\frac{m}{s}$

The car stops, so its final speed is zero. Solving for a:

$a=\frac{v_0^2}{2x}\\a=-\frac{(10.83\frac{m}{s})^2}{2(1.7*10^{-2}m)}\\a=-3449.67\frac{m}{s^2}$