Question

students a and b are standing inside a train that is accelerating to the right with a magnitude of 2 m/s². student a throws a ball to the left toward student b with a horizontal velocity of magnitude 4 m/s relative to the train. at the instant the ball leaves student a’s hand, the speed of the train relative to the ground is 10 m/s to the right. the ball takes 0.5 second to reach student b. immediately before the ball reaches student b, the horizontal speed of the ball relative to the ground is
a 7 m/s
b 6 m/s
c 5 m/s

Explanation:

Step1: Analyze train's velocity change

The train is accelerating at $a = 2\ m/s^{2}$ for $t=0.5\ s$. Using the formula $v = v_0+at$, where $v_0 = 10\ m/s$, $a = 2\ m/s^{2}$ and $t = 0.5\ s$. The velocity of the train after $0.5\ s$ is $v_{train}=10 + 2\times0.5=10 + 1=11\ m/s$ to the right.

Step2: Analyze ball's velocity relative to train

The ball has a velocity of $v_{ball - train}=- 4\ m/s$ (negative because it is thrown to the left relative to the train).

Step3: Calculate ball's velocity relative to ground

Using the relative - velocity formula $v_{ball - ground}=v_{ball - train}+v_{train}$. Substitute the values: $v_{ball - ground}=-4 + 11 = 7\ m/s$.

Answer:

A. 7 m/s