Question

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?

Answer 1

The speed of the train after an additional 5 s has elapsed is 9.45 m/s

What is acceleration?

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

How to determine the acceleration in 5.4 s

• 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

How to determine the final velocity after 5 s

• 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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