Question

An automobile and train move together along parallel paths at 23.2 m/s. The automobile then undergoes a uniform acceleration of −4 m/s 2 because of a red light and comes to rest. It remains at rest for 22.7 s, then accelerates back to a speed of 23.2 m/s at a rate of 2.09 m/s 2 . How far behind the train is the automobile when it reaches the speed of 23.2 m/s, assuming that the train speed has remained at 23.2 m/s? Answer in units of m.

Answer 1

Answer:

Train will 588.215 m ahead of automobile

Explanation:

We have given initially speed of both train and automobile is same as 23 m/sec

After that acceleration of automobile $a=-4m/sec^2$

Initial velocity u = 23 m/sec

Final velocity v = 0 ( as it comes to rest )

According to first law of motion

v=u+at

$0=23-4\times t$

t = 5.75 sec

Now according to third law of motion $v^2=u^2+2as$

$0^2=23^2-2\times 4\times s$

s = 66.125 m

Now automobile accelerates at with acceleration $a=2.09m/sec^2$

Initial velocity u = 23 m/sec

Final velocity v = 23.2 m/sec

So $time\ t =\frac{v-u}{a}=\frac{23.2-23}{2.09}=0.0956sec$

Distance traveled by automobile in 0.0956 sec

$s=ut+\frac{1}{2}at^2=23\times 0.0956+\frac{1}{2}\times 2.09\times 0.0956^2=2.208m$

So total distance traveled by automobile = 66.125+2.208 = 68.333 m

Total time = 5.75 +22.7+0.0956=28.545 sec

Sp distance traveled by train in 28.545 sec

= 28.545×23 = 656.548 m

So distance between automobile and train = 656.548 - 68.33=588.215 m

So train will 588.215 m ahead of automobile