Question

A driver traveling at 100 km/hr notices the speed limit changes to 50 km/hr when he is entering town. He takes 0.9 second to slow down to 50 km/hr with constant acceleration. We want to know how many meters the driver travels while slowing down.


50 POINTS TO THE PERSON WHO GETS IT RIGHT FIRST!!!!
HURRRY PLEASE!

Answer 1

Answer:

B) $\Delta x = (\frac{v+v_0}{2})t$

Explanation:

In order to find the distance covered by the driver while slowing down, we can use the following suvat equation:

$\Delta x = (\frac{v+v_0}{2})t$

where

$\Delta x$ is the distance covered

v is the final velocity

$v_0$ is the initial velocity

t is the time

For the car in this problem,

$v_0 = 100 km/h \cdot \frac{1000}{3600}=27.8 m/s$

$v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s$

t = 0.9 s

Substituting,

$\Delta x = (\frac{13.9+27.8}{2})(0.9)=18.8 m$