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 1.75 s and the second in 1.17 s. Find the constant acceleration of the train.
I know how to find average acceleration (vcf- vxi/tf-ti)but don't understand how to find CONSTANT acceleration.
acceleration= deltav/time=(8.6/1.17 -8.6/1.75)/time
now time here is (1.75+1.17)/2
solve for acceleration.
To find the constant acceleration of the train, you can use the equations of motion. Let's break down the problem step-by-step:
Step 1: Find the initial velocity of the train (v₀).
Since the train is already departing when Liz arrives, it means its initial velocity is zero.
v₀ = 0 m/s
Step 2: Find the final velocity of the train (v).
To find the final velocity of the train, we need to determine the distance it travels in the given time intervals and divide it by the respective time.
For the first car:
v₁ = (distance₁) / (time₁)
= (8.60 m) / (1.75 s)
For the second car:
v₂ = (distance₂) / (time₂)
= (8.60 m) / (1.17 s)
Step 3: Calculate the acceleration (a).
The acceleration can be determined using the equation:
a = (v - v₀) / t
For the first car:
a₁ = (v₁ - v₀) / t₁
= (v₁ - 0) / (1.75 s)
For the second car:
a₂ = (v₂ - v₀) / t₂
= (v₂ - 0) / (1.17 s)
Step 4: Calculate the average acceleration (aavg).
To find the constant acceleration, we can take the average of the two acceleration values calculated in Step 3.
aavg = (a₁ + a₂) / 2
Substitute the values found in Steps 2 and 3 into the equation to compute the average acceleration.
To find the constant acceleration of the train, you can use the equations of motion derived from the definition of acceleration. The equation that you will need is the one that relates initial velocity (vi), final velocity (vf), acceleration (a), and displacement (d):
vf^2 = vi^2 + 2ad
Let's solve this step-by-step:
Step 1: Determine the initial and final velocities.
Since Liz is watching the cars go by, the initial velocity of the train relative to Liz is zero. Therefore, vi = 0 m/s.
Step 2: Calculate the displacement.
Given that each car is 8.60 m long, the displacement (d) is equal to the length of the car. Therefore, d = 8.60 m.
Step 3: Calculate the final velocity of the first car.
The final velocity (vf1) of the first car can be calculated using the formula:
vf1 = vi + a*t
where t is the time it takes for the car to pass Liz. From the problem, t1 = 1.75 s. Since vi = 0, we can simplify the equation to:
vf1 = a*t1
Step 4: Calculate the final velocity of the second car.
Follow the same procedure as step 3, but this time use the time given for the second car: t2 = 1.17 s.
vf2 = a*t2
Step 5: Use the equation of motion to find acceleration.
Now, we have two equations:
vf1^2 = 2ad
vf2^2 = 2ad
Since the same acceleration applies to both cars, we can equate the two equations:
a*t1^2 = a*t2^2
Divide both sides by 'a':
t1^2 = t2^2
Since the acceleration term cancels out, we are left with:
t1 = t2
Comparing the given times, we see that t1 = 1.75 s and t2 = 1.17 s. Since they are not equal, something must be amiss. Please double-check the given values and re-calculate if necessary.
If both times are equal, it means the acceleration is constant. To find the acceleration, substitute one of the final velocities into the equation:
a = (vf1^2 - vi^2) / (2d)
Remember that vi = 0 m/s and vf1 is the final velocity of the first car. Plug in these values and the displacement to find the constant acceleration.