Question

I am so lost and feel completely stupid because I don't understand this!!....

An object moves for 3 s with a constant velocity of 4 m/s, then moves for 2s with a constant velocity of -2 m/s. The object then stops for 1s and then moves with a constant velocity of -5m/s for 5s.

How fast would the object have to travel at constant speed to get back to its starting point in .5 s?

Answer 1

Break it down into the stages given:

Travel at 4m/s for 3 s : distance travelled = ?
Travel at -2m/s for 2s : Distance travelled = ? (Note the minus sign means in opposite direction)
Travel at -5 m/s for 5 s : Distance travelled = ? (again note the direction)
(During the 1s stop - no motion - so you can ignore it)
So the total distance the object is now from where it started is the sum of the 3 distances above (NOTE the direction)

SO It now has to travel that distance to get 'home'
Distance/time = speed . You know the distance, you know the time, = work out the speed.
AT ALL TIMES be careful of the direction (i.e. the minus signs!)

Answer 2

Thank You. For my final answer I got 97 m/s . But, if by being "careful" you mean to just leave the final negative answer as a positive one, then i did that.

Answer 3

I don't agree with your answer
1st phase +12m
2nd phase - 4m
3rd phase -24m

Net result = -16m

So object is -16m from home - so has to travel +16m to get there
To do this in 0.5s means it has to travel +32m/s

Answer 4

my first phase i got : +18 m.....
i got different answers than you because i did the distance formula --- d= (.5)(a)(t^2)
so my first phase would have been (.5)(4)(3^2) = 18....

is this formula wrong?

Answer 5

distance = speed x time ..never mind i understand.

Answer 6

yes- that is the formula for constant acceleration - where a is the acceleration
s = ut + 0.5 (at^2)
In this case a=0 - constant speed

so s = ut (distance - velocity x time

I notice an error in MY answer

Third phase - 25m
so Net distance = -17m
Return to home = +34m/s

Answer 7

Ok. Got it ! (:

would you mind if i asked you part b of this question? It wants to know "how far away from its starting point is the object after 11s?"

Do I multiply 34 m/s and 11 s?

Answer 8

I think that you have already answered this:
Phase 1 - 3s
Phase 2 - 2s
Pause - 1s
Phase 3 = 5s

Total = 11s

Answer 9

So in 3+2+1+5 (11),
the object has traveled 17 m/s.

Answer 10

I don't think so - and anyhow - the question is not 'how far has it travelled' but - 'how far is it from starting point'
Also - you said 34m/s - but a distance is measured in m

Answer 11

If you walk 2 miles along a straight road, then turn rtound and walk a mile back, then oyu have travelled 3 miles, but are only 1 mile from starting point.

This is where the minus signs are important - they indicate the direction of travel.

Answer 12

so...

forwards (12m) and
backwards (-4m) would put you at 16 m traveled and backwards another (-25m) would be 33 m for the distance from the starting point??

Answer 13

NO
Forward 12
Backward 4 (i.e. -4)
Backward 25 (i.e. -25)
1s wait
+12-4-25 = -17

Answer 14

Ok. i completely understand. I really appreciate your help!

Answer 15

I am no longer lost nor do i feel stupid (:

Answer 16

:-)