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. ThanksNugatory 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.
It is a constant acceleration problem. Read https://www.ncl.ac.uk/webtemplate/a...mechanics/kinematics/equations-of-motion.html.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
Alternatively, try the Khan Academy. It seems you need a course in physics, not just a bit of help: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 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
That's why I recommend Khan Academy. That's a reliable source for SUVAT.Nugatory said:Google for “SUVAT equations “. They relate speed, distance, and acceleration.
Thanks so much! That got me where I needed to be.Nugatory said:Google for “SUVAT equations “. They relate speed, distance, and acceleration.
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 said:Alternatively, try the Khan Academy. It seems you need a course in physics, not just a bit of help:
https://www.khanacademy.org/science/physics
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.
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.
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.
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.
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.