a=(v_(f)-v_(i))/(t) s=(1)/(2)(v_(f)+v_(i))t s=v_(i)t+(1)/(2)at^2 2as=v_(f)^2-v_(i)^2 the acceleration if it takes 160 s for a train to go from 28m/s east to 12m/s west?

Question

a=(v_(f)-v_(i))/(t) s=(1)/(2)(v_(f)+v_(i))t s=v_(i)t+(1)/(2)at^2 2as=v_(f)^2-v_(i)^2 the acceleration if it takes 160 s for a train to go from 28m/s east to 12m/s west?

Answer 1

Here's how to calculate the acceleration:1. **Understand the directions:** "East" and "West" indicate opposite directions. We need to account for this by assigning one direction as positive and the other as negative. Let's consider East as positive and West as negative.2. **Define variables:*** *vᵢ* (initial velocity) = 28 m/s (East, so positive)* *v꜀* (final velocity) = -12 m/s (West, so negative)* *t* (time) = 160 s* *a* (acceleration) = what we want to find3. **Choose the right equation:** The first equation of motion, *a* = (*v꜀* - *vᵢ*) / *t* , is the most suitable for this problem.4. **Substitute and solve:***a* = (-12 m/s - 28 m/s) / 160 s*a* = (-40 m/s) / 160 s*a* = -0.25 m/s²5. **Interpret the result:** The acceleration is -0.25 m/s². The negative sign indicates the acceleration is directed westward, which makes sense since the train is slowing down in the eastward direction and eventually reverses to travel westward.**Answer:** The acceleration of the train is -0.25 m/s².

Question

a=(v_(f)-v_(i))/(t) s=(1)/(2)(v_(f)+v_(i))t s=v_(i)t+(1)/(2)at^2 2as=v_(f)^2-v_(i)^2 the acceleration if it takes 160 s for a train to go from 28m/s east to 12m/s west?

Solution

Answer