(1) Before the typhoon landed in Xiamen
Obviously, s shadow =300.
∫△ t = t1-T2 5: 00 pm on September 30th to15: 30 am on October 2nd, 10 * * 35.5 hours.
∴△t=35.5h
The speed is the same before and after landing on Lotus Island.
The speed after landing on Lianhua Island can be regarded as the speed before landing on Lianhua Island.
∴v0=△s÷△t=940km÷35.5h≈26.48km/h
According to S=vt, the distance between s and gate is -v0t+340km =-26.48t+340 (0
(2) After the typhoon landed in Xiamen,
According to S=vt, s distance gate =vtt= 18t.
The average hourly radius of the affected area is reduced by 4km.
∴S Shadow = -4t+300
∴ To sum up, -26.48t+340 (0
S distance gate =
18t(t & gt； 12.84 )
300(0 & lt； t & lt 12.84)
S shadow =
-4t+300(t & gt； 12.84 )
When s left the door
∴S distance gate -s shadow < 0
-26.48t+340-300 & lt； 0(0 & lt； t & lt 12.84)
18t-(-4t+300)& lt； 0(t & gt； 12.84 )
The solution is 1.5 1
∴ 13.64- 1.5 1= 12. 13≈ 12h7min
A: It affects Xiamen about 12 hours and 7 minutes.
(2) Solution: Suppose that the distance (km) between S and Chang represents the distance between Typhoon Longwang and Typhoon Nanchang; Vt (km/h) indicates the speed after the typhoon landed in Xiamen.
Drawings:
∵CD⊥AB
∴Rt△CDB Rt△ADC
∴CD2+BD2=BC2 AD2+CD2=AC2
CB = vtt
∴CB= 18t
∴cd=bc×sin∠cbd= 18t×sin 15
BD = BC×cos∠CBD = 18t×cos 15
∴AD=500- 18t×cos 15
∴ac=√(500- 18t×cos 15)2+( 18t×sin 15)2
Distance from constant = √ (500-18t× cos15) 2+(18t× sin15) 2
The average hourly radius of the affected area is reduced by 4km.
∴S Shadow = -4t+300
When you leave Chang.
Shadow from Zhang zhen < 0