Finding the average velocity of a dropped bolt

[Guest]

A construction worker drops a bolt while on a high rise buildng 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s(t)= 320 - 5t^2 and 0 < t < 8.

b) Find the average velocity for the interval 3 < t < 8

avg velocity: delta d/delta t

. . .= [s(8) - s(3)] / [8 - 3]

. . .= [(320 - 5(8)²) - (320 - 5(3)²)] / 5

. . .= (320 - 320 - 320 + 45) / 5

. . .= -275/5

. . .= -55

I got -55, but its suppose to be +55. What did I do wrong? The only thing I can think of is that delta t is suppose to be 3 - 8, instead of 8 - 3, but then should it be s(3) - s(8), if you switch them around?

Thank you for your help.
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Edited by stapel -- Reason for edit: formatting

 

[pka]

Re: Finding avg velocity

I got -55, but its suppose to be +55, what did I do wrong,

Why is that the case? v(t)=s'(t) which is negative.

 

[TchrWill]

A construction worker drops a bolt while on a high rise buildng 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s(t)= 320 - 5t^2 and 0 < t < 8.

b) Find the average velocity for the interval 3 < t < 8

avg velocity: delta d/delta t

. . .= [s(8) - s(3)] / [8 - 3]

. . .= [(320 - 5(8)²) - (320 - 5(3)²)] / 5

. . .= (320 - 320 - 320 + 45) / 5

. . .= -275/5

. . .= -55

I got -55, but its suppose to be +55. What did I do wrong? The only thing I can think of is that delta t is suppose to be 3 - 8, instead of 8 - 3, but then should it be s(3) - s(8), if you switch them around?

Thank you for your help.

The distance fallen in 3 seconds is 5(3)^2 = 45m.

The distance fallen in 8 seconds is 5(8)^2 = 320m

The average speed between 3 and 8 seconds is (320 - 45)/5 = 55m/s

 

[pka]

TchrWill, You do know that there is difference between velocity and speed!
Velocity is v(t)=s’(t) and speed(t)=|v(t)|, they are not the same.
The posted question is about velocity not speed.

 

[skeeter]

pka is correct.

velocity is a vector, having both magnitude and direction ... -55 m/s is a velocity, the (-) sign indicates direction (down).

speed is a scalar and is the magnitude of velocity ... 55 m/s is the speed.

the original question did ask for average velocity.

Math types need to be very careful when they pose their physics problems.

 

[TchrWill]

TchrWill, You do know that there is difference between velocity and speed!
Velocity is v(t)=s’(t) and speed(t)=|v(t)|, they are not the same.
The posted question is about velocity not speed.

True, the velocity of a body is a vector quantity having the same magnitude as its speed, but including the direction of motion. Thus, in stating the velocity of a body, both the speed and direction of motion must be included. Since the direction of the dropped object is vertical (its direction), both the average velocity and average speed are 55m/s.

 

[daon]

Actually, depending on which direction you call positive, it could be +55m/s or -55m/s. Its all relative.

It is somewhat standard to call downward ("towards earth") negative, so your -55m/s should be the correct answer.