For which of the following position-time (x-t) graphs is acceleration zero?

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ScanSolve AI Tutor · Accepted Answer

Step 1: Acceleration is zero when velocity is constant. In an x-t graph, constant velocity appears as a straight line (non-zero slope). Observe the graphs: Graph (A) is a straight line (constant slope), so acceleration is zero throughout. Graphs (B), (C), (D) are curved or piecewise, indicating changing velocity. Answer for Q12: (A) Step 1: Maximum velocity from v-t graph is the highest v value (peak ordinate). Step 2: Maximum displacement is the maximum area under v-t curve up to that point (max ∫ v dt). Step 3: Maximum acceleration is the maximum slope of v-t graph (max (Δ v)/(Δ t)). Step 4: Average displacement is total area under curve divided by time interval (1/T ∫₀^T v dt). (Specific values require graph scales; read from given v-t graphs for A, B, C, D.) The area under v-t graph represents displacement (∫ v dt = Δ x), not velocity or acceleration. Velocity = ordinate value. Acceleration = slope. neither (A) nor (B); it is displacement

For which of the following position-time (x-t) graphs is acceleration zero?

Step-by-step solution

Acceleration is zero when velocity is constant.

Maximum velocity from $v$ - $t$ graph is the highest $v$ value (peak ordinate).

Maximum displacement is the maximum area under $v$ - $t$ curve up to that point ( $\max \int v \, dt$ ).

Maximum acceleration is the maximum slope of $v$ - $t$ graph ( $\max \frac{\Delta v}{\Delta t}$ ).

Average displacement is total area under curve divided by time interval ( $\frac{1}{T} \int_0^T v \, dt$ ).

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For which of the following position-time (x-t) graphs is acceleration zero?

Step-by-step solution

Acceleration is zero when velocity is constant.

Maximum velocity from vvv-ttt graph is the highest vvv value (peak ordinate).

Maximum displacement is the maximum area under vvv-ttt curve up to that point (max⁡∫v dt\max \int v \, dtmax∫vdt).

Maximum acceleration is the maximum slope of vvv-ttt graph (max⁡ΔvΔt\max \frac{\Delta v}{\Delta t}maxΔtΔv​).

Average displacement is total area under curve divided by time interval (1T∫0Tv dt\frac{1}{T} \int_0^T v \, dtT1​∫0T​vdt).

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More Physics Questions

Maximum velocity from $v$ - $t$ graph is the highest $v$ value (peak ordinate).

Maximum displacement is the maximum area under $v$ - $t$ curve up to that point ( $\max \int v \, dt$ ).

Maximum acceleration is the maximum slope of $v$ - $t$ graph ( $\max \frac{\Delta v}{\Delta t}$ ).

Average displacement is total area under curve divided by time interval ( $\frac{1}{T} \int_0^T v \, dt$ ).