A block of ice slides down a chute with an acceleration of 18 fps

[SurnameLong]

A block of ice slides down a chute with an acceleration of 18 feet per second per second. (a) If it is going 6 feet per second at a certain instant, how far does it slide in the next 3 seconds? (b) How far from the first position has it gone by the time its speed is 72 feet per second?

I was given a table with feet per seconds, approximate value of g is 32.

dv/dt=-32, v=ds/dt=-32t + A, s= -16t²+At +B
I thought 18 should've been subbed in for A, 6 feet per second for v, and 3 for t, yet, this yields 84.
I tried adding 18 to -32 for dv/dt because of the per second per second notation. Incorrect answer.
The answer is 99 ft for (a) and 143 feet for (b). I have no idea as to how they got these answers.

 

[Deleted member 4993]

A block of ice slides down a chute with an acceleration of 18 feet per second per second. (a) If it is going 6 feet per second at a certain instant, how far does it slide in the next 3 seconds? (b) How far from the first position has it gone by the time its speed is 72 feet per second?

I was given a table with feet per seconds, approximate value of g is 32.

dv/dt=-32, v=ds/dt=-32t + A, s= -16t²+At +B
I thought 18 should've been subbed in for A, 6 feet per second for v, and 3 for t, yet, this yields 84.
I tried adding 18 to -32 for dv/dt because of the per second per second notation. Incorrect answer.
The answer is 99 ft for (a) and 143 feet for (b). I have no idea as to how they got these answers.

That is your t= 0.

So v(0) = 6

 

[Ishuda]

A block of ice slides down a chute with an acceleration of 18 feet per second per second. (a) If it is going 6 feet per second at a certain instant, how far does it slide in the next 3 seconds? (b) How far from the first position has it gone by the time its speed is 72 feet per second?

I was given a table with feet per seconds, approximate value of g is 32.

dv/dt=-32, v=ds/dt=-32t + A, s= -16t²+At +B
I thought 18 should've been subbed in for A, 6 feet per second for v, and 3 for t, yet, this yields 84.
I tried adding 18 to -32 for dv/dt because of the per second per second notation. Incorrect answer.
The answer is 99 ft for (a) and 143 feet for (b). I have no idea as to how they got these answers.

You need to pick a reference time for when the block is going 6 ft per second and a reference distance at that time. Both the reference time and distance can be chosen as zero. That is we will measure elapsed time and elapsed distance from when and where the block is when it is going 6 feet per second.

So, neglecting units,
\(\displaystyle d(t)\, =\, \underset{0}{\overset{t}{\int}}\, v(t)\, dt\)
\(\displaystyle v(t)\, =\, \underset{0}{\overset{t}{\int}}\, a(t)\, dt\)
\(\displaystyle a(t)\, =\, 18\)
\(\displaystyle v(0)\, =\,\, 6\)
\(\displaystyle d(0)\, =\,\, 0\)