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until a train is a safe distance from the station it must travel at 5 m…

Question

until a train is a safe distance from the station it must travel at 5 m/s. once the train is on open track it can speed up to 45 m/s. if it takes a train 8 seconds to reach 45 m/s, what is the acceleration of the train? (round your answer to the nearest whole number.)
4 m/s²
5 m/s²
6 m/s²
7 m/s²

Explanation:

Step1: Recall the acceleration formula

Acceleration \( a \) is calculated by the change in velocity \( \Delta v \) divided by the time taken \( \Delta t \), so \( a=\frac{\Delta v}{\Delta t} \).

Step2: Determine initial and final velocities

Initial velocity \( v_i = 5 \, \text{m/s} \), final velocity \( v_f = 45 \, \text{m/s} \). So \( \Delta v = v_f - v_i = 45 - 5 = 40 \, \text{m/s} \).

Step3: Calculate acceleration

Time \( \Delta t = 8 \, \text{s} \). Then \( a = \frac{40}{8} = 5 \, \text{m/s}^2 \).

Answer:

B. \( 5 \, \text{m/s}^2 \)