Initially stationary, a train has a constant acceleration of 0.8 m/s2. (a) What is its speed after 27 s? m/s (b) What is the total time required for the train to reach a speed of 41 m/s?

Given:

By using Kinematics equation, we get

→

(a)

At t = 27 seconds,

→

(b)

When  

→

Thus the response above is right.

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(a) v1 = 21.6 m/s

(b) t = 51.25 s

Explanation:

Use kinematics equation

v1 = v0 + at

Given

v0 = 0 = initial velocity

a = 0.8 m/s^2 = acceleration

(a) t = 27 seconds

v1 = v0 + at = 0 + 0.8*27 = 21.6 m/s

(b) v1 = 41 m/s

v1 = v0 + at

solve for t

t = (v1-v0)/a = (41-0)/0.8 = 51.25 s

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000