Liz rushes down onto a subway platform to find her train already departing. She stops and watches the cars go by. Each car is 8.60 m long. The first moves past her in 1.70 s and the second in 1.52 s. Find the constant acceleration of the train.
A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are d1 = 4.00 m to the north, d2 = 1.90 m northeast, and d3 = 1.00 m at θ = 30.0° west of south (figure below). Starting at the same initial point, an expert golfer could make the hole in what single displacement?
To find the constant acceleration of the train, we can use the equation:
\(a = \frac{{(v_f - v_i)}}{t}\)
Where:
- \(a\) is the acceleration
- \(v_f\) is the final velocity
- \(v_i\) is the initial velocity
- \(t\) is the time taken
In this case, the initial velocity, \(v_i\), is the velocity of the train right before it passed Liz, and the final velocity, \(v_f\), is the velocity of the train after it passed Liz.
Given:
- Length of each car, \(d = 8.60 m\)
- Time taken for the first car to pass, \(t_1 = 1.70 s\)
- Time taken for the second car to pass, \(t_2 = 1.52 s\)
We can find the initial and final velocities by dividing the distance traveled by each car by the time taken for them to pass Liz.
For the first car:
\(v_{i1} = \frac{d}{t_1}\)
For the second car:
\(v_{i2} = \frac{d}{t_2}\)
Now we can calculate the acceleration using the equation mentioned earlier:
\(a = \frac{{(v_{f2} - v_{i1})}}{t_2}\)
Let's substitute the values and calculate the acceleration.
To find the constant acceleration of the train, we can use the equation of motion for uniformly accelerated linear motion:
v = u + at,
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time taken.
Let's break down the information given in the problem:
- Each car is 8.60 m long.
- The time taken for the first car to move past Liz is 1.70 s.
- The time taken for the second car to move past Liz is 1.52 s.
First, let's calculate the initial and final velocities of the train for both cars.
For the first car:
Initial velocity, u1 = 0 m/s (since the car starts from rest)
Final velocity, v1 = (distance traveled) / (time taken) = 8.60 m / 1.70 s
For the second car:
Initial velocity, u2 = v1 (since the first car is already in motion)
Final velocity, v2 = (distance traveled) / (time taken) = 8.60 m / 1.52 s
Now that we have the initial and final velocities, we can find the acceleration using the equation v = u + at and rearranging it as:
a = (v - u) / t.
For the first car:
Acceleration, a1 = (v1 - u1) / t1 = v1 / t1
For the second car:
Acceleration, a2 = (v2 - u2) / t2 = (v2 - v1) / t2
Now, substitute the values and calculate the accelerations:
For the first car:
a1 = v1 / t1 = (8.60 m / 1.70 s) / 1.70 s
For the second car:
a2 = (v2 - v1) / t2 = ((8.60 m / 1.52 s) - (8.60 m / 1.70 s)) / 1.52 s
By evaluating the above expressions, you can find the constant acceleration of the train.