Answer to Question #224632 in Mechanical Engineering for Papi Chulo

Question

A car of length 5 m is being driven parallel alongside a 100 m long train at the same

speed. The front of the car is 10 m behind the rear of the train. At the instant the train

decelerates at 0.4 ms² the car accelerates at 0,15 ms². How long will it take for the rear

of the car to pass the front of the train?

Answer 1

Let $S_{10}=5+10+100=115(m), S_{20}=0$

$v_{01}=v_{02}=v_{0}$

$a_1=-0.4\ m/s^2, a_2=0.15\ m/s^2$

The equation of the motion of a train

$$S_1(t)=S_{10}+v_0t+\dfrac{a_1t^2}{2}$$

$$S_2(t)=S_{20}+v_0t+\dfrac{a_2t^2}{2}$$

The rear of the car pass the front of the train

$$S_1=S_2$$

$$S_{10}+v_0t+\dfrac{a_1t^2}{2}=S_{20}+v_0t+\dfrac{a_2t^2}{2}$$

$$t=\sqrt{\dfrac{2(S_{10}-S_{20})}{a_2-a_1}}$$

$$t=\sqrt{\dfrac{2(115\ m-0\ m)}{0.15\ m/s^2-(-0.4\ m/s^2)}}\approx20.45\ s$$

It will take 20.45 seconds for the rear of the car to pass the front of the train.

We cannot define  the distance the truck and the train will cover in this time because the initial velocity of car and train are not given.

If $v_{01}=v_{02}=v_{0}\ m/s,$ then

the distance the train will cover in this time is

$$L_1(20.45)=(v_0(20.45)-\dfrac{0.4(20.45)^2}{2})\ m$$

$$=(20.45v_0-83.636) m$$

the distance the truck will cover in this time

$$L_2(20.45)=(v_0(20.45)+\dfrac{0.15(20.45)^2}{2}) \ m$$

$$=(115+v_0(20.45)-\dfrac{0.4(20.45)^2}{2})\ m$$

$$=(20.45v_0+31.364)\ m$$