Primeira página
/
Matemática
/
When an Alvia high -speed train that is stopped in Barcelona leaves the station, its speed increases (accelerates) at a constant rate of 3 ,888 kilometers per hour squared ((km)/(hr^2)) . If his the time, in hours, it takes for the train to reach a speed of 250(km)/(hr) , which of the following equations best describes this situation?

Pergunta

When an Alvia high -speed train that is stopped in Barcelona leaves the
station, its speed increases (accelerates) at a constant rate of 3 ,888
kilometers per hour squared ((km)/(hr^2)) . If his the time, in hours, it takes for
the train to reach a speed of 250(km)/(hr) , which of the following equations
best describes this situation?

When an Alvia high -speed train that is stopped in Barcelona leaves the station, its speed increases (accelerates) at a constant rate of 3 ,888 kilometers per hour squared ((km)/(hr^2)) . If his the time, in hours, it takes for the train to reach a speed of 250(km)/(hr) , which of the following equations best describes this situation?

Solução

expert verifiedVerification of experts
4.3201 Voting
avatar
TatianiProfissional · Tutor por 6 anos

Responder

To find the equation that best describes the situation, we need to use the formula for acceleration, which is given by:<br /><br />\[ a = \frac{\Delta v}{\Delta t} \]<br /><br />where:<br />- \( a \) is the acceleration,<br />- \( \Delta v \) is the change in velocity,<br />- \( \Delta t \) is the change in time.<br /><br />Given:<br />- The acceleration \( a \) is \( 3888 \frac{km}{hr^2} \),<br />- The initial velocity \( v_0 \) is \( 0 \frac{km}{hr} \) (since the train starts from rest),<br />- The final velocity \( v \) is \( 250 \frac{km}{hr} \).<br /><br />We need to find the time \( t \) it takes for the train to reach the speed of \( 250 \frac{km}{hr} \).<br /><br />Using the formula for acceleration, we can rearrange it to solve for \( \Delta t \):<br /><br />\[ \Delta t = \frac{\Delta v}{a} \]<br /><br />Substituting the given values:<br /><br />\[ \Delta t = \frac{250 \frac{km}{hr} - 0 \frac{km}{hr}}{3888 \frac{km}{hr^2}} \]<br /><br />\[ \Delta t = \frac{250}{3888} \]<br /><br />\[ \Delta t \approx 0.0643 \text{ hours} \]<br /><br />Therefore, the equation that best describes this situation is:<br /><br />\[ t = \frac{v}{a} \]<br /><br />where:<br />- \( t \) is the time in hours,<br />- \( v \) is the velocity in \( \frac{km}{hr} \),<br />- \( a \) is the acceleration in \( \frac{km}{hr^2} \).<br /><br />So, the equation that best describes this situation is:<br /><br />\[ t = \frac{v}{3888} \]
Clique para avaliar: