Tewtor.ai

What do you already know about how position–time graphs relate to velocity and acceleration?

Key idea to use here:

• The horizontal axis is not time $t$ but $t^2$. The graph shows $x$ versus $t^2$ and it’s a straight line with negative slope. So we can write $x = a + b\,t^2,$ with constant $a$ and constant slope $b<0$.

From this:

• Velocity is $v = \dfrac{dx}{dt} = 2bt.$
• Acceleration is $a_x = \dfrac{dv}{dt} = 2b,$ a constant (and negative since $b<0$).

Interpretation:

• The car’s acceleration is constant and negative (points opposite the chosen positive direction).
• Starting “from rest” at $t=0$ matches $v(0)=0$.
• Because $v=2bt$ becomes more negative with time, the car speeds up in the negative direction at a constant rate.

Which choice matches that? “The car speeds up at a constant rate” (choice D). If the positive direction were chosen along the initial motion, the constant acceleration would be positive; either way, the magnitude of acceleration is constant, so the speed increases linearly in time.

Quick check: Why not constant speed? If $x$ vs $t^2$ is linear, $x$ vs $t$ is quadratic, which means non-constant velocity.

What was the key insight here? That a straight line in $x$ vs $t^2$ means $x \propto t^2$, giving constant acceleration.