20 (part 1 of 2)
The initial speed of a body is 6.61 m/s.
What is its speed after 3.92 s if it accelerates
uniformly at 3.46 m/s/s?
Answer in units of m/s
21 (part 2 of 2) 10.0 points
What is its speed after 3.92 s if it accelerates
uniformly at −3.46 m/s/s?
Answer in units of m/s
What is the answer for 21?
Initial speed plus the acceleration times the time.
6.61 m/s + 3.46 m/s/s times 3.92 s
6.61 m/s + ____ m/s (fillin the blank with 3.46 times 3.92).
The same principle applies, but you will be multiplyint by -3.46 which will result in a decrease in the speed.
-6.95
To find the speed of the body after a certain time when it accelerates uniformly, you can use the equation of motion:
Final speed (V) = initial speed (U) + acceleration (a) * time (t)
For the first question:
1. Write down the given values:
- Initial speed (U) = 6.61 m/s
- Acceleration (a) = 3.46 m/s²
- Time (t) = 3.92 s
2. Plug the values into the formula:
V = U + a * t
V = 6.61 m/s + 3.46 m/s² * 3.92 s
3. Calculate the value:
V = 6.61 m/s + 13.5792 m/s
V ≈ 20.1892 m/s
So, the speed of the body after 3.92 s with an acceleration of 3.46 m/s² is approximately 20.1892 m/s.
For the second question:
1. Again, write down the given values:
- Initial speed (U) = 6.61 m/s
- Acceleration (a) = -3.46 m/s² (negative value indicates deceleration)
- Time (t) = 3.92 s
2. Plug the values into the formula:
V = U + a * t
V = 6.61 m/s + (-3.46 m/s²) * 3.92 s
3. Calculate the value:
V = 6.61 m/s - 13.5792 m/s
V ≈ -6.9692 m/s
So, the speed of the body after 3.92 s with an acceleration of -3.46 m/s² is approximately -6.9692 m/s. Note that the negative sign indicates the direction of deceleration.