12. From observing the graph, we can see that from t=0s to t=3s, the graph has the same slope. If we can find that slope then, that value will be the acceleration at any time in between, since
From the graph, in between 0s and 3s,
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13. Between 3s and 6s, the curve is flat, ie there is no acceleration and the velocity is constant which also means that we have a uniform motion in our time interval, the displacement during this time interval is given by the equation:
is the time interval, here v=6m/s
Hence,
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14. a) As we have seen in the previous case, the acceleration is zero in the time interval t=3s to t=6s, since the slope in this time interval is zero
b)The acceleration is positive when the slope is positive (ie, the velocity is increasing with time), the time periods with the said attributes are, t=0s to t=3s and t=6s to t=8s
c) The acceleration is negative when the slope is negative (ie, the velocity is decreasing with time), the time periods with the said attributes is, t=8s to t=10s
Use the graph below for questions 11 15. Velocity versus Time 10 123 45678910 Time (s)...
9. Write the result of (4.36 x10-1-3.6x 10-2) (6.50x 10+7.94x 10°) in scientific notation with the correct significant figures 10. If Samantha ran a distance of 10 m in a direction 30° north of east in 5 s, and then, a distance of 6 m southwest 2 s later, what is her average velocity in rectangular form (or component form)? Use the graph below for questions 11 - 15. Velocity versus Time 10 1 2 3 45678910 Time (s) 12....
Please answer both based on the graph 13. What is the displacement from t 3 s tot 6 s? 14. For what time interval(s) is the acceleration zero? positive? negative? a./ Use the graph below for questions 11 15. Velocity versus Time 10 3 123 456 78 9 10 Time (s)
Please use the graph to answer 11-14 9. Write the result of (4.36 x 101-36x 10(6.50x 10+7.94 x 10) in scientific notation with the correct significant figures 10. If Samantha ran a distanice of 10 m in a direction 30P north of east in 5 s, and then, a distance of 6 m southwest 2 s later, what is ber average velocity in rectangular form (or component form)? Use the graph below for questions 11 - 15. Velocity versus Time...
Use the graph below for questions 12 15 Velocity versus Time 10 1 23 456 789 10 Time (s) 2. What is the average acceleration of the particle fromt2s tot9s? 3. What is the instantaneous acceleration at t 5 s? what is the displacem ent from t = 4 s to 1 = 8 s? 14. 5. Sketch the acceleration versus time graph derived from the velocity versus time graph. Acceleration versus Time 3 -3 1 23 45 678 9...
9. Write the result of (4.36 x 101-36x 10(6.50x 10+7.94 x 10) in scientific notation with the correct significant figures 10. If Samantha ran a distanice of 10 m in a direction 30P north of east in 5 s, and then, a distance of 6 m southwest 2 s later, what is ber average velocity in rectangular form (or component form)? Use the graph below for questions 11 - 15. Velocity versus Time 10 E 7 26 1 23 456789...
Use the graph below for questions 11 1 Velocity versus Thne to 12845TS t Time () 14. Find the displacenient from t 0 to t = 6 15 Sketch the acceleration versus time graph derived from the velocity versus time graph Acceleration versus Time 0 -3 -4 1 2 3 4 5 6 7 89 10 Time (s) Extra Credit
Please answer 12-13 based on the graph. Use the graph below for questions 12- 15. Velocity versus Time 10 1 2 3 45678 9 10 Time (s) 12. What is the average acceleration of the particle from t 2s to t9s? 13, what is the instantaneous acceleration at 5s?
Extra Credit Calculate the velocity of the particle from t-Ostot 16s whose acceleration versus time graph is Acceleration versus Time a(m/s2 14 12 10 t(s) 123 456 78910 11 12 13 14 15 16
Use the graph below for questions 12 15 Velocity versus Time 1 2 3 4 5 6 78 9 10 Time (s) 12, what is the average acceleration of the particle from t = 2 s to t-9 s? 13. What is the instantaneous acceleration at t 5s? G. If a projectile is launched with an initial velocity of 20 m/s from a height of 50 m in th Mehort ho ng dos it take the projectile to return to...