# Distance traveled when decreasing velocity

• brslagle
brslagle
Homework Statement
An amusement park ride launches a rider at an
angle of 90 degrees to the horizontal, with an initial
velocity of 50 m/s. Ignoring air resistance, what
will be the rider’s height at t = 1.5, t = 4 and t = 6
seconds?
Relevant Equations
unsure
dont know where to start. Other than it will take 5 seconds for v = 0m/s

What equations have been mentioned in class or your textbook so far? Look at them, think about what they mean, one or more of them will be relevant.

Nugatory said:
What equations have been mentioned in class or your textbook so far? Look at them, think about what they mean, one or more of them will be relevant.
I don’t have any to go off of. I’m reviewing physics (haven’t had it in 5 years) and am stuck on this type of problem, so additional help would be appreciated. Thanks

brslagle said:
Homework Statement: An amusement park ride launches a rider at an
angle of 90 degrees to the horizontal, with an initial
velocity of 50 m/s. Ignoring air resistance, what
will be the rider’s height at t = 1.5, t = 4 and t = 6
seconds?
Relevant Equations: unsure

dont know where to start. Other than it will take 5 seconds for v = 0m/s
It is a constant acceleration problem. Read https://www.ncl.ac.uk/webtemplate/a...mechanics/kinematics/equations-of-motion.html.

brslagle said:
I don’t have any to go off of. I’m reviewing physics (haven’t had it in 5 years) and am stuck on this type of problem, so additional help would be appreciated. Thanks
Alternatively, try the Khan Academy. It seems you need a course in physics, not just a bit of help:

brslagle
brslagle said:
I don’t have any to go off of. I’m reviewing physics (haven’t had it in 5 years) and am stuck on this type of problem, so additional help would be appreciated. Thanks
Google for “SUVAT equations “. They relate speed, distance, and acceleration.

brslagle
Nugatory said:
Google for “SUVAT equations “. They relate speed, distance, and acceleration.
That's why I recommend Khan Academy. That's a reliable source for SUVAT.

Nugatory
Nugatory said:
Google for “SUVAT equations “. They relate speed, distance, and acceleration.
Thanks so much! That got me where I needed to be.

PeroK said:
Alternatively, try the Khan Academy. It seems you need a course in physics, not just a bit of help:

Thanks. The SUVAT problems were the only questions I was having issues on, so I dont think an entire course is necessary, but thanks for the suggestion.

PeroK

## What is the formula to calculate the distance traveled when velocity decreases uniformly?

The formula to calculate the distance traveled when velocity decreases uniformly (constant deceleration) is given by: $$d = \frac{(v_i + v_f)}{2} \times t$$, where $$v_i$$ is the initial velocity, $$v_f$$ is the final velocity, and $$t$$ is the time taken.

## How do you determine the deceleration if the distance and initial and final velocities are known?

You can determine the deceleration using the formula: $$a = \frac{(v_f^2 - v_i^2)}{2d}$$, where $$v_i$$ is the initial velocity, $$v_f$$ is the final velocity, and $$d$$ is the distance traveled.

## What role does time play in calculating the distance traveled during deceleration?

Time is a crucial factor in calculating the distance traveled because it helps determine how long the deceleration occurs. The distance can be calculated using the formula: $$d = v_i \times t - \frac{1}{2} a \times t^2$$, where $$v_i$$ is the initial velocity, $$a$$ is the deceleration, and $$t$$ is the time.

## Can distance traveled be negative when velocity is decreasing?

No, distance traveled cannot be negative. Distance is a scalar quantity and always represents the magnitude of the path traveled, which is always a positive value, even if the velocity is decreasing.

## How does initial velocity affect the distance traveled when decelerating to a stop?

The initial velocity directly affects the distance traveled when decelerating to a stop. A higher initial velocity means that the object will travel a greater distance before coming to a stop, given the same rate of deceleration. The relationship can be expressed as: $$d = \frac{v_i^2}{2a}$$, where $$v_i$$ is the initial velocity and $$a$$ is the deceleration.

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