Question

A car has a velocity of 49.29 m/s. It then accelerates at a uniform rate of -4.00 m/s² for the next 5.40 seconds. what distance does it travel?

Answer 1

$\begin{gathered} v_o=49.29\text{ m/s} \\ a=-4.00m/s^2 \\ t=5.40\text{ s} \\ x_o=0\text{ m} \\ x_f=? \\ x_f=x_o+v_ot+\frac{at^2}{2} \\ x_f=0\text{ m+}(49.29\text{ m/s})(5.40\text{ s})+\frac{(-4.00m/s^2)(5.40\text{ s})^2}{2} \\ x_f=\text{207}.85m \\ It\text{ travels 207.85 m} \end{gathered}$