velocity v(t)=2t^2-6t+5 integer time slowing down

Key concept: Use the acceleration $a(t)=v'(t)$ and compare its sign to the velocity. A particle slows down exactly when those signs are opposite.

Step by step

We have

$v(t)=2t^2-6t+5.$

A particle slows down when velocity and acceleration have opposite signs.

Step 1: Find acceleration

$a(t)=v'(t)=4t-6.$

Step 2: Test positive integers

• At $t=1$: $v(1)=1>0, \qquad a(1)=-2<0$ Opposite signs, so the particle is slowing down.

• At $t=2$: $v(2)=1>0, \qquad a(2)=2>0$ Same sign, so it is not slowing down.

So the integer time is

$\boxed{1}.$

Pitfall alert

Do not confuse slowing down with negative velocity. A particle can move in the positive direction and still slow down if acceleration is negative.

Try different conditions

If the question asked for the time when the particle is speeding up, then you would look for velocity and acceleration having the same sign instead.

Further reading

speeding up, derivative, sign of acceleration