A train,starting from rest,accelerates along the platform at a uniform rate of 0.6 m/s².A passenger standing on the platform is 5 m away from the door when the train starts to pull away and heads toward the door at an acceleration of 1.2 m/s².How long does it take the passenger to reach the door? A)3.0 s B)17 s C)7.1 s D)5.6 s E)4.1 s

Question

Quizplus · Accepted Answer

First, let's find the velocity of the train when the passenger starts moving. We can use the equation v = u + a*t, where u is initial velocity (0 m/s for the train), a is acceleration (0.6 m/s^2 for the train), and t is time.

v = u + a*t
v = 0 + 0.6*t
v = 0.6t

After time t, the velocity of the train will be 0.6t m/s. At this point, the passenger starts moving with an acceleration of 1.2 m/s^2. We can use the same equation to find the time it takes for the passenger to reach the door.

Let's call the distance between the passenger and the door x. We know that x = 5 m. We want to find the time t2 it takes for the passenger to travel this distance with an acceleration of 1.2 m/s^2.

x = ut + 0.5*a*t^2
5 = 0 + 0.5*1.2*t2^2
5 = 0.6*t2^2
t2^2 = 8.3...
t2 = sqrt(8.3) = 2.88...

Therefore, the time it takes for the passenger to reach the door is approximately 2.88 seconds, or 7.1 seconds to one decimal place. The answer is C).

A Train,starting from Rest,accelerates Along the Platform at a Uniform