Question

A certain car takes 30m to stop when it is traveling at 25m/s. If a pedestrian is 28m in front of this car when the driver starts braking (starting at 25m/s), how long does the pedestrian have to get out of the way?

Answer 1

Answer:

It take 1.78033 second get away

Explanation:

We have given that a car takes 30 m to stop when its speed is 25 m/sec

As the car stops its final speed v = 0 m/sec

Initial speed u = 25 m/sec

Distance s = 30 m

From third law of motion $v^2=u^2+2as$

So $0^2=25^2+2\times a\times 30$

$a=-10.4166m/sec^2$

Now in second case distance s = 28 m

So $v^2=25^2+2\times -10.4166\times 28$

$v^2=41.666$

v = 6.4549 m/sec

Now from first equation of motion v=u+at

So $6.4549=25-10.4166\times t$

t = 1.78033 sec