# Acceleration and velocity question

• bango
In summary, the question is asking for the time it would take for a cart on a similar ramp to reach a velocity of 90km/h. The data provided is in the form of a graph with velocity (cm/s) on the y-axis and time (s) on the x-axis. To answer the first question, the equation v = (166cm/s^2)t is used, where v is velocity and t is time. To convert the given velocity of 90km/h to cm/s, unit conversion is used. For the second question, the value of t is simply plugged into the equation to find the corresponding velocity. The slope of the graph can also be used to extrapolate the graph to longer times and higher velocities
bango
hey, i have a lab due for a grade 10 physics class tomorrow. I am not too smart with this stuff, and I'm stuck with this question. can someone please help me out;

Question: how long would it take a cart on a similar ramp to reach a velocity of 90km/h?

i did a lab where we measured the speeds of a cart going down a ramp with a ticker tape timer thing.

i calculated that the average velocity was 166cm/s [down]

... there's one more question too;

calculate the velocity the cart would have if it was allowed to run down a similar ramp for a period of 5.0s

im sure this is real basic stuff, I am just stuck

my data was

time 0.0 velocity (cm/s) 0
time 0.05 velocity 16
time 0.15 velocity 31
time 0.25 velocity 47
time 0.35 velocity 63
time 0.45 velocity 72
time 0.55 velocity 99
time o.65 velocity 107
time 0.75 velocity 123
time 0.85 velocity 183
time 0.95 velocity 154

its basically uniform when plotted on a velocity vs time graph

Your graph has all the information you need. Draw the best straight line you can through your data and extrapolate (extend) your graph to longer times and higher velocities if needed. For the first question, you need to convert from km/hr to cm/sec. At what time does the graph cross 90km/hr? (Average velocity is not the most direct approach to answer this question. The number you have for average velocity is not what the data says. Are you sure that is velocity?)

For the seond question, what velocity is on your graph at t = 5 sec?

Extrapolation of the graph can be done in your mind if not done on paper. What is the slope of the graph? How can you use that to decide where points would be at larger times and larger velocities?

Try and get an equation for the graph and everything should be easy from there.

im still stuck, i understand what you're saying. but i don't know how to do it. i don't know proper equations, that's why I am on these forums

bango said:
im still stuck, i understand what you're saying. but i don't know how to do it. i don't know proper equations, that's why I am on these forums

Your data graph is "uniform" you said. I take that to mean a straight line. The general equation of a line is

y = mx + b

In your case y is velocity, and x is time. b is the value of y when x is zero, which is zero for you. The equation that represents your line is

v = mt

and you calculated the slope to be 166m/s^2, except that you did not get the units right. We call the change in velocity divided by the change in time acceleration, and that is the slope of your graph. Your equation is

v = (166cm/s^2)t

Using that equation, you can figure out what time gives a velocity of 90km/hr, but you need that velocity expressed in cm/sec. The best way to convert units is to multiply what you have by fractions that are equal to one that replace unwanted units with wanted units.

$$90\frac{km}{hr}=90\frac{km}{hr}\left[\frac{1000m}{km}\right]\left[\frac{100cm}{m}\right]\left[\frac{1hr}{60min}\right]\left[\frac{1min}{60sec}\right]= how-many\left[\frac{cm}{sec}\right]$$

Plug the result into the equation relating v and t and solve for t.

For the last question, plug in for t to find v.

## 1. What is the difference between acceleration and velocity?

Acceleration is the rate at which an object's velocity changes over time, while velocity is the rate of change of an object's position over time. In simpler terms, acceleration is a measure of how fast an object is speeding up or slowing down, while velocity is a measure of how fast and in what direction an object is moving.

## 2. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. This can be represented by the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 3. What are the units of acceleration and velocity?

The SI unit for acceleration is meters per second squared (m/s²), while the SI unit for velocity is meters per second (m/s). In some cases, acceleration may also be measured in units of gravity (g), where 1 g is equivalent to 9.8 m/s².

## 4. How does acceleration affect an object's motion?

Acceleration directly affects an object's motion by changing its velocity. If an object experiences a positive acceleration, it will speed up in the direction of the acceleration. If an object experiences a negative acceleration (also known as deceleration), it will slow down in the opposite direction of the acceleration. If an object experiences no acceleration, its velocity will remain constant.

## 5. Can an object have a constant velocity and changing acceleration?

Yes, an object can have a constant velocity and changing acceleration. This is because velocity only describes an object's speed and direction of motion, while acceleration describes the change in velocity over time. So, an object can have a constant velocity in one direction while experiencing changing acceleration in a different direction.

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