To find the change in velocity from time $t = 1$ second to $t = 5$ seconds, we need to use the relationship between acceleration and velocity. The change in velocity ($\Delta v$) can be found by integrating the acceleration over the given time interval:

$\Delta v = \int_{t_1}^{t_2} a(t) \, dt$

Given:

• From $t = 0$ s to $t = 4$ s, acceleration $a(t) = 2.0 \, \text{m/s}^2$.
• From $t = 4$ s to $t = 7$ s, acceleration $a(t) = -3.0 \, \text{m/s}^2$.

We need to find the change in velocity from $t = 1$ second to $t = 5$ seconds.

Steps:

1. Calculate the change in velocity from $t = 1$ s to $t = 4$ s: $\Delta v_1 = \int_{1}^{4} 2.0 \, dt = 2.0 \times (4 - 1) = 2.0 \times 3 = 6.0 \, \text{m/s}$

2. Calculate the change in velocity from $t = 4$ s to $t = 5$ s: $\Delta v_2 = \int_{4}^{5} -3.0 \, dt = -3.0 \times (5 - 4) = -3.0 \times 1 = -3.0 \, \text{m/s}$

3. Add the changes in velocity: $\Delta v = \Delta v_1 + \Delta v_2 = 6.0 \, \text{m/s} + (-3.0 \, \text{m/s}) = 3.0 \, \text{m/s}$

Answer:

The change in velocity from $t = 1$ second to $t = 5$ seconds is 3.0 m/s.

Would you like more details, or do you have any other questions?

Related Questions:

1. How do you calculate displacement from a velocity-time graph?
2. What is the significance of the area under the curve in acceleration-time graphs?
3. Can you explain the relationship between acceleration, velocity, and displacement?
4. How do initial conditions affect the calculation of velocity from acceleration?
5. How would the problem change if acceleration were a function of time rather than a constant?

Tip:

When analyzing motion using acceleration-time graphs, remember that the area under the curve between two time intervals gives you the change in velocity over that period.