Questions: displacement u v · t a · t → 1/2 a · t^2 time t a · t → v · t = Δx = v · t + 1/2 a · t^2

Transcript text: \begin{aligned} &\text{displacement} \\ &u \\ &v \cdot t \\ &\text{a} \cdot t \\ &\rightarrow \frac{1}{2} \text{a} \cdot t^2 \\ &\text{time t} \\ &\text{a} \cdot t \\ &\rightarrow v \cdot t \\ &= \Delta x = v \cdot t + \frac{1}{2} \text{a} \cdot t^2 \end{aligned}

Solution

Solution Steps

The image depicts a derivation of the displacement equation for constant acceleration.

Step 1: Area under the velocity-time graph represents displacement

The shaded area under the velocity vs. time graph represents the displacement (change in position). This area can be divided into a rectangle and a triangle.

Step 2: Area of the rectangle

The area of the rectangle is base times height, which corresponds to _v_ * _t_ where _v_ is the initial velocity and _t_ is the time.

Step 3: Area of the triangle

The area of the triangle is 1/2 * base * height. The base is _t_ (time). The height is the change in velocity, which is equal to _at_ (acceleration \* time). Thus, the triangle's area is 1/2 * _t_ * _at_ = 1/2 * _a_ * _t_² .