Question

A 3.2 km train is moving at 87% the speed of light on a new train track. It is approaching a tunnel that is 1 km long. From the perspective of a stationary observer, can the train just fit inside the tunnel completely the moment the nose is about to come out the other side? If not, what fraction of the speed of light would it need to be moving in order to accomplish this? (use 3 sig figs for your final answers)

a. No, 0.950

b. No, 0.990

c. No, 0.820

d. yes

No, 0.820

d.

Yes

Answer 1

The actual length of the bou en l = 3.2 kesan The length of the laund l = 1 kma 1610 m = 3.2 x 103 m Here, bet el-3x10 m is) be els velocity of the tra light. So the vebeuty of the tram es. 8577 of c 87 % of a sh V 2 87 gotte v. – 0.870 with speed مل As the train moving near to light, so according to spread theory of adetvety, the longth of the sot train will be contracted with respect an oven ovserver at rest Led l be the contracted length of the train. Now, if a l'zl, then to the bian well fit en the tunnel Nom from spilength contraction of speaal theory dhe relativety,

fur l'= 1 (0.87€ = 3.976? [-0.87'] l = 1557.76 So l'he, fille length of the tunnel) So, I will not fit en. When l'ali then the train well & just fd in inside the tunnel. Let to for then, the webcity of the trans in A gown from the formule of length conte contraction, 0 [ [r="/Wir na [here ... al[in]? (-)" li 1- 12 12

1_ l? Cr d. c - 5 1 - le ste = 0.950 - Then in the sequerd saction. Bu Ivi og goe - The e lo requaed velaty So the correct option Opbeon a. No, 0.990
The answer is in the images. Page number is mentioned at right top.