As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.4 s. if the train's acceleration remains constant
, what is its speed after an additional 5.0 s has elapsed?
1 answer:
The speed of the train after an additional 5 s has elapsed is 9.45 m/s
<h3>What is acceleration? </h3>
This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
- a is the acceleration
- v is the final velocity
- u is the initial velocity
- t is the time
<h3>How to determine the acceleration in 5.4 s</h3>
- Initial velocity (u) = 0 m/s
- Final velocity (v) = 4.9 m/s
- Time (t) = 5.4 s
- Acceleration (a) =?
a = (v – u) / t
a = (4.9 – 0) / 5.4
a = 4.9 / 5.4
a = 0.91 m/s²
<h3>How to determine the final velocity after 5 s</h3>
- Initial velocity (u) = 4.9 m/s
- Acceleration (a) = 0.91 m/s²
- Time (t) = 5 s
- Final velocity (v) = ?
a = (v – u) / t
0.91 = (v – 4.9) / 5
Cross multiply
v – 4.9 = 0.91 × 5
v – 4.9 = 4.55
Collect like terms
v = 4.55 + 4.9
v = 9.45 m/s
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