Here's what I did but I don't know another way of solving this without using point slope form.

(0, 6)

(1.5, y)

y-y1=m(x-x1)

y-6=-2(1.5-0)

y-6= -3

y=3m/s

velocity =3m/s forward

Here's what I did but I don't know another way of solving this without using point slope form.

(0, 6)

(1.5, y)

y-y1=m(x-x1)

y-6=-2(1.5-0)

y-6= -3

y=3m/s

velocity =3m/s forward

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The simplest way to solve this problem is by using the basic acceleration formula V= at Where V is velocity, a is acceleration, and t= time. This problem gives an initial velocity an acceleration and a time asking for an ending velocity. We can find the ending velocity two different ways. We can take the acceleration of -2m/s times the time of 1.5 seconds, and find the change in velocity of -2*1.5= -3m/s then we add the initial velocity to the change in velocity 6m/s +-3m/s = 3m/s or you can solve the equation using the slope line intercept formula.

(0, 6)

(1.5, y)

y-y1=m(x-x1)

y-6=-2(1.5-0)

y-6= -3

y=3m/s

velocity =3m/s forward

You got the correct answer, Mathalina!

Physically, the constant, negative 2m/s/s acceleration means that the velocity will negate 2 m/s each second ... v= 6m/s to start ... 4m/s after 1s ... 2m/s at t=2s ... 0 at t=3s ... -2m/s (backward now) at t=4s ... -4m/s at t=5s ... much like heading uphill, slowing to a stop, then reversing downhill faster & faster :)

_{0}+a*t, where v_{0} is the initial velocity, t is the time elapsed. Brad explained a physical meaning of this, thanks to him.

6.0m/s - (2.0m/s^{2})(1.5s) = 6.0m/s - 3.0m/s =** 3.0m/s**

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