Question

c) the velocity of a particle for ( 0 leq t leq 9 ) is given in the graph above. at which of the following values of ( t ) is the particle speeding up?
( t = 3 )
( t = 4 )
( t = 5 )
( t = 7 )

Explanation:

Step1: Recall the condition for speeding up

A particle is speeding up when \(v(t)\) and \(a(t)\) (the derivative of \(v(t)\)) have the same sign. The slope of the velocity - time graph gives the acceleration \(a(t)=\frac{dv}{dt}\).

Step2: Analyze \(t = 3\)

At \(t = 3\), \(v(3)>0\) (velocity is positive). The slope of the \(v(t)\) graph (acceleration) at \(t = 3\) is negative (\(a(3)<0\)). Since \(v(3)\) and \(a(3)\) have opposite signs, the particle is not speeding up at \(t = 3\).

Step3: Analyze \(t = 4\)

At \(t = 4\), \(v(4)=0\). When \(v(t) = 0\), the particle is momentarily at rest (not speeding up or slowing down in the sense of non - zero speed change).

Step4: Analyze \(t = 5\)

At \(t = 5\), \(v(5)<0\) (velocity is negative). The slope of the \(v(t)\) graph (acceleration) at \(t = 5\) is negative (\(a(5)<0\)). Since \(v(5)\) and \(a(5)\) have the same sign, the particle is speeding up at \(t = 5\).

Step5: Analyze \(t = 7\)

At \(t = 7\), \(v(7)<0\) (velocity is negative). The slope of the \(v(t)\) graph (acceleration) at \(t = 7\) is positive (\(a(7)>0\)). Since \(v(7)\) and \(a(7)\) have opposite signs, the particle is not speeding up at \(t = 7\).

Answer:

\(t = 5\)