Question

the accompanying figure shows the velocity v = ds/dt = f(t) (m/sec) of a body moving along a coordinate line. a. when does the body reverse direction? b. when is it moving at a constant speed? c. graph the bodys speed for 0 ≤ t ≤ 10. d. graph the acceleration, where defined. choose the correct graph of the acceleration, where defined.

Explanation:

Step1: Determine direction - change

The body reverses direction when velocity changes sign. From the velocity - time graph, velocity changes sign at \(t = 2\) s and \(t = 7\) s.

Step2: Identify constant - speed intervals

The body is moving at a constant speed when the magnitude of the velocity is constant. This occurs on the intervals \([0,1]\), \([3,4]\), \([5,6]\), \([7,8]\) and \([9,10]\).

Step3: Graph the speed

Speed is the absolute - value of velocity. For \(0\leq t\leq1\), \(v(t)\) is increasing from \(0\) to \(5\), so \(s(t)=v(t)\). For \(1 < t\leq2\), \(v(t)\) is decreasing from \(5\) to \(0\), so \(s(t)=v(t)\). For \(2 < t\leq3\), \(v(t)\) is negative and increasing from \( - 5\) to \(0\), so \(s(t)=-v(t)\). For \(3 < t\leq4\), \(v(t)\) is positive and constant at \(5\), so \(s(t) = 5\). For \(4 < t\leq5\), \(v(t)\) is positive and decreasing from \(5\) to \(0\), so \(s(t)=v(t)\). For \(5 < t\leq6\), \(v(t)\) is positive and constant at \(5\), so \(s(t)=5\). For \(6 < t\leq7\), \(v(t)\) is positive and decreasing from \(5\) to \(0\), so \(s(t)=v(t)\). For \(7 < t\leq8\), \(v(t)\) is negative and constant at \(-5\), so \(s(t)=5\). For \(8 < t\leq9\), \(v(t)\) is negative and increasing from \(-5\) to \(0\), so \(s(t)=-v(t)\). For \(9 < t\leq10\), \(v(t)\) is negative and constant at \(-5\), so \(s(t)=5\).

Step4: Calculate acceleration

Acceleration \(a(t)=\frac{dv}{dt}\). On \([0,1]\), \(a(t) = 5\) (since \(v(t)\) changes from \(0\) to \(5\) in \(1\) second). On \([1,2]\), \(a(t)=-5\) (since \(v(t)\) changes from \(5\) to \(0\) in \(1\) second). On \([2,3]\), \(a(t)=5\) (since \(v(t)\) changes from \(-5\) to \(0\) in \(1\) second). On \([3,4]\), \(a(t) = 0\) (constant velocity). On \([4,5]\), \(a(t)=-5\) (since \(v(t)\) changes from \(5\) to \(0\) in \(1\) second). On \([5,6]\), \(a(t)=0\) (constant velocity). On \([6,7]\), \(a(t)=-5\) (since \(v(t)\) changes from \(5\) to \(0\) in \(1\) second). On \([7,8]\), \(a(t)=0\) (constant velocity). On \([8,9]\), \(a(t)=5\) (since \(v(t)\) changes from \(-5\) to \(0\) in \(1\) second). On \([9,10]\), \(a(t)=0\) (constant velocity). The correct acceleration - graph is the one that has constant non - zero values during the intervals of non - zero slope of the velocity graph and zero values during the intervals of zero slope of the velocity graph.

Answer:

a. The body reverses direction at \(t = 2\) s and \(t = 7\) s.
b. It is moving at a constant speed on the intervals \([0,1]\), \([3,4]\), \([5,6]\), \([7,8]\) and \([9,10]\).
c. (Graph the speed as described in Step 3).
d. (The correct acceleration graph is the one that follows the acceleration values calculated in Step 4. Without seeing the exact details of the graphs A, B, C, D in a clear way, we can't choose a specific letter - option, but the acceleration graph should have non - zero constant values during the intervals where the velocity graph has a non - zero slope and zero values during the intervals where the velocity graph has a zero slope).