📝 Position vs time graph analysis calculus (24 MCQs)

📖 Explanation: The slope of the s‑versus‑t curve at any instant equals the instantaneous velocity of the particle. This follows directly from the definition of derivative: $v(t)=\frac{ds}{dt}$. Hence a steeper slope indicates a larger speed, while a flat slope means zero velocity.

📖 Explanation: Concavity reflects the sign of the second derivative s&#039;&#039;(t), which is the acceleration. When the curve is concave up (s&#039;&#039;(t)>0), the acceleration is positive; when concave down (s&#039;&#039;(t)<0), the acceleration is negative.

📖 Explanation: A positive slope means the velocity is positive, while a concave down shape indicates a negative acceleration. Since velocity and acceleration have opposite signs, the particle’s speed is decreasing, i.e., it is slowing down during that interval.

📖 Explanation: A horizontal tangent means the instantaneous velocity is zero. Concave up indicates that the second derivative s&#039;&#039;(t) is positive, so the acceleration is positive. Thus the particle is momentarily at rest but experiencing a positive acceleration that will cause it to start moving forward.

📖 Explanation: For $0\le t\le2$, $v=2t>0$ and $a=2>0$; velocity and acceleration share the same sign, so the speed increases. On $2\le t\le4$, $v= -2(t-2)<0$ while $a=-2<0$; again the signs match, but the speed decreases because the magnitude of velocity diminishes. Hence only the first interval shows speeding up.

📖 Explanation: Since acceleration equals the second derivative, the graph that is concave up has a positive second derivative, while the concave‑down graph has a negative second derivative of the same magnitude only if the curvature is symmetric. Typically, a concave‑up shape implies a larger positive acceleration magnitude than the negative magnitude of the other.

📖 Explanation: When the slope (velocity) is positive, the particle moves forward. As s&#039;&#039;(t) switches from positive to negative, the acceleration changes from adding to the velocity to subtracting from it. Consequently, the particle transitions from increasing speed (speeding up) to decreasing speed (slowing down).}

📖 Explanation: A positive slope indicates a positive velocity. Concave up means the second derivative s&#039;&#039;(t) is positive, so the acceleration is also positive. Therefore both velocity and acceleration are positive, meaning the particle’s speed is increasing in the positive direction.

📖 Explanation: Crossing the origin with a positive slope implies the derivative $ds/dt$ is positive at that instant. Hence the particle’s velocity is positive as it passes through the origin, indicating motion in the positive direction.

📖 Explanation: Speed decreases when the velocity and acceleration have opposite signs. A positive velocity combined with a negative acceleration reduces the magnitude of the velocity over time, so the particle is slowing down despite moving in the positive direction.

📖 Explanation: A horizontal tangent gives $v=0$. Concave up indicates a positive acceleration (s&#039;&#039;>0). A positive acceleration acting on a particle at rest will cause it to acquire a positive velocity, so the particle will begin moving in the positive direction.

📖 Explanation: A positive slope indicates a positive velocity. Concave down means the second derivative s&#039;&#039;(t) is negative, so the acceleration is negative. Thus the particle is moving forward while decelerating.

📖 Explanation: $v(t)=3kt^{2}$ and $a(t)=6kt$. For $t>0$, both $v$ and $a$ are positive only if $k>0$; then velocity and acceleration share the same sign, guaranteeing the speed increases.

📖 Explanation: A momentary stop occurs when the tangent is horizontal ($v=0$) but the curvature is non‑zero, so acceleration is non‑zero and the particle will resume motion. Being at rest for an interval requires both velocity and acceleration to be zero throughout that interval.

📖 Explanation: Constant acceleration implies a constant second derivative, leading to a quadratic dependence of position on time. The graph of a uniformly accelerated particle is a parabola opening upward (if acceleration is positive) or downward (if negative).

📖 Explanation: A horizontal tangent means the derivative $ds/dt$ equals zero at that point, which directly corresponds to zero instantaneous velocity. It does not imply a maximum or minimum displacement unless additional context about surrounding points is given.

📖 Explanation: Velocity is the first derivative (slope) of the position‑time graph, while acceleration is the second derivative (curvature). A horizontal tangent gives zero slope (zero velocity) while a non‑zero curvature yields a non‑zero second derivative, meaning acceleration can be present even when the particle is momentarily at rest.

📖 Explanation: Constant positive acceleration means $a$ is a positive constant, so integrating twice yields $s(t)=\frac{1}{2}at^{2}+vt_{0}+s_{0}$, a quadratic function opening upward. Therefore the position‑time graph is a concave‑up parabola for all future times.

📖 Explanation: The sign of the position function indicates the side of the origin. A negative value of $s(t)$ means the particle lies to the left (or negative side) of the origin on the $s$-axis.

📖 Explanation: When the slope is positive, velocity is positive. Concave up (s&#039;&#039;>0) gives a positive acceleration, while concave down (s&#039;&#039;<0) gives a negative acceleration. Thus the first case has acceleration positive, the second case negative.

📖 Explanation: Traveling east corresponds to motion in the positive direction, so the graph must have a positive slope. Steady speeding up means acceleration is positive, giving a concave‑up shape. Hence the graph is positively sloped and concave up.

📖 Explanation: During $2<t<4$ the velocity remains positive while the speed decreases, indicating that acceleration opposes the motion. Therefore the acceleration must be negative on that interval.

📖 Explanation: For $0\le t\le1$, velocity $v=1$ and acceleration $a=0$; the particle moves at constant speed. For $1\le t\le3$, $v=2(t-1)$ and $a=2$. Speed increases until $v$ reaches a maximum at the endpoint $t=3$; however, the transition from speeding up to slowing down occurs when acceleration changes sign, which happens at $t=2$ where the curvature of the graph switches from linear to quadratic.

📖 Explanation: When velocity and acceleration share the same sign, the acceleration contributes to increasing the magnitude of the velocity, leading to a larger speed. If the signs differ, the acceleration works against the motion, reducing speed. Therefore identical signs are necessary for a particle to speed up.