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 acceleration of 1.2 m/s².How far away from the door is the passenger after 3 seconds? A)2.7 m B)5.4 m C)0.40 m D)2.3 m E)0.90 m

Question

Quizplus · Accepted Answer

We can use the following kinematic equations to solve the problem:

v = u + at
s = ut + 1/2 at^2

Where:
u = initial velocity (0 m/s)
v = final velocity
a = acceleration
t = time
s = distance

First, we can find the final velocity of the train after 3 seconds:

v = u + at
v = 0 + 0.6 m/s^2 × 3 s
v = 1.8 m/s

Next, we can use this velocity to find the distance traveled by the train:

s = ut + 1/2 at^2
s = 0 × 3 s + 1/2 × 0.6 m/s^2 × (3 s)^2
s = 2.7 m

Finally, we can find the distance between the passenger and the door by subtracting the distance traveled by the train from the initial distance:

distance = 5 m - 2.7 m
distance = 2.3 m

Therefore, the correct answer is D (2.3 m).

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