Hyperloop: Speed

s = 560 km (560000 m)

t = 0.5 h (1800s)

v = s/t

v = 560000 m / 1800s

v = 311.111 m/s

Converting units, we have a distance of $560 \; \text{km} \cdot \dfrac{1000 \; \text{m}}{1 \; \text{km}} = 560000 \; \text{m}$ and a time of $0.5 \; \text{hr} \cdot \dfrac{60 \; \text{min}}{1 \;\text{hr}} \cdot \dfrac{60 \;\text{sec}}{1\; \text{min}} = 1800 \; \text{sec}.$ Therefore, the speed of the Hyperloop is $\dfrac{560000}{1800} \approx \boxed{311.11}\; \text{m/s}.$

A lógica é simples. Comece transformando as unidades de comprimento e de tempo para metro e segundos respectivamente, em seguida aplique a fórmula para velocidade média. 560 KM = 560 * 1000 m = 560.000m 30 mim = 30 * 60 s = 1800s Velocidade média = 560000 / 1800 = 311.11 m/s

$v=\frac{s}{t}$

$v=\frac{560000}{1800}$

$v=\boxed{311.111}$

The speed of the Hyperloop is $560 km/h \times 2 = 1120 km/h$ .So the speed in $m/s$ is $\frac{1120 \times 1000m}{3600 s}$ = $\frac{11200m}{36 s}$ = $311 \frac{1}{9} m/s$

using speed formula s=d/t=560000/1800=311.11m/s

speed=distance/time; speed=560 1000/30 60; 311.11m/s

{560 x 1000 [m]} : {30 x 60 [s]} = 560000/1800 = 311.110 [m/s] . Answer : 3.1111E+2

Let $d$ be the distance traveled. Let $t$ be the time to travel the distance, $d$ . Let $v$ be the speed. Since $v$ is a constant, we have $d = vt \implies v = \dfrac {d}{t}.$ For $d = \text {560 km}$ and $t = \text {0.5 hr} = 1800 \, \text{s}$ , we have $v = \boxed {311.11 \, \text{m/s}}.$

560 KM is 560,000 in meters. You divide that by 30 to get the speed in minutes and divide that by 60 to get it in 60 which makes it the answer 311.11 repeating

Speed = Distance/Time. Therefore Speed = 560/(1/2)= 1120 km/hr. To convert km/hr into m/sec 1120*5/18=311.11

the Hyperloop will cover distance 560 km i.e, 560000 m in time 30 min i.e, 1800 sec. Speed = distance/time u = 560000/1800 u = 311.11 m/s

560 km=560000 m half an hour=30 minutes=1800 seconds s=d/t =560000/1800 =311.11