A train is moving along a straight line with a constant acceleration ‘a’. A girl standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 600 to the horizontal. The girl has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s2, is
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We have a train moving with constant acceleration 'a'. A girl standing in the train throws a ball forward with speed 10 meters per second at an angle of 60 degrees to the horizontal. She needs to move 1.15 meters forward inside the train to catch the ball at the same height. We need to find the train's acceleration.
To solve this problem, we analyze the motion from the train's reference frame. Since the train is accelerating, this is a non-inertial frame. In this frame, the ball experiences two accelerations: gravity acting downward with magnitude g, and a pseudo-force acting backward with magnitude a. The initial velocity has horizontal component 5 meters per second and vertical component 5 root 3 meters per second.
To find the time of flight, we use the vertical motion equation. Since the ball returns to the same height, the vertical displacement is zero. This gives us the equation: zero equals 5 root 3 times t minus g over 2 times t squared. Solving for t, we get t equals 10 root 3 over g. Using g equals 10 meters per second squared, the time of flight is root 3 seconds.
Now we analyze the horizontal motion. The girl moves 1.15 meters forward to catch the ball, which means the ball's horizontal displacement relative to the train is 1.15 meters. Using the horizontal motion equation with initial velocity 5 meters per second and pseudo-acceleration negative a, we substitute the time of flight root 3 seconds to get the equation: 1.15 equals 5 root 3 minus 3a over 2.
Now we solve for the acceleration. From our equation, a equals two-thirds times the quantity 5 root 3 minus 1.15. Using the approximation root 3 equals 1.73, we calculate: a equals two-thirds times 8.65 minus 1.15, which equals two-thirds times 7.5, giving us a final answer of 5 meters per second squared. This matches our expected result and satisfies all the given conditions in the problem.