Time() 0 2 4 6 8 10 12 14 16
Velocity() 0 0 2 5 10 15 20 22 22
By measuring the slope of your curve, find the magnitude of the instantaneous acceleration at times = 9s
i don't know how to solve it
Perhaps they want you to draw a graph of V(t), with a smooth line through the points, and measure the slope of the tangent to that curve at t = 9s.
The acceleration rate is not constant for your set of numbers, so there will be some uncertainty based upon how you draw the curve.
A very good estimate of the acceleration at t = 9s can be obtained by dividing the velocity change between t = 8 and 10 s by the time interval (2 s)..
a(t=9) = (15 - 10)/2 = 7.5 m/s^2
i did the graph already and i put the anwser 7.5m/s^2 and i got it wrong
i got it thank
Sorry my mistake. Right approach; math/typing error. I meant 2.5 m/s^2
To find the magnitude of the instantaneous acceleration at a specific time, we can use the concept of differentiation. The acceleration of an object can be determined by finding the rate of change of its velocity with respect to time. In other words, we need to compute the derivative of the velocity function.
Given the data points for time and velocity, we can approximate the derivative using a method called finite differences. First, we find the difference in velocity between consecutive time intervals. Then, we divide that difference by the duration of the time interval.
Here is how you can approach solving the problem step by step:
Step 1: Construct a table for the time and velocity values:
```
Time(s): 0 2 4 6 8 10 12 14 16
Velocity: 0 0 2 5 10 15 20 22 22
```
Step 2: Calculate the differences in velocity between consecutive time intervals:
```
Time(s): 0 2 4 6 8 10 12 14 16
Velocity: 0 0 2 5 10 15 20 22 22
Difference: 0 2 3 5 5 2 2
```
Step 3: Calculate the average acceleration for each interval (using finite differences):
```
Time(s): 0 2 4 6 8 10 12 14 16
Velocity: 0 0 2 5 10 15 20 22 22
Difference: 0 2 3 5 5 2 2
Acceleration: 0 1 1 1 0 -3 0
```
Step 4: Approximate the instantaneous acceleration at time t = 9s by interpolating between adjacent acceleration values. Since the time interval between t = 8s and t = 10s is 2 seconds, we can assume the acceleration is constant for that interval. Therefore, the acceleration at t = 9s is equal to the acceleration at t = 8s, which is 0.
Therefore, the magnitude of the instantaneous acceleration at t = 9s is 0.