# Determine the distance travelled by the mass while it is accelerating upwards.?

I don't know how to answer this physics question, basically there is a graph, velocity against time graph.
it increases rapidly until 0.7 seconds (2.1 m/s)and starts to level off after that and is a straight line (2.5m/s) after 1 second. It sarts to decrease by levelling off a bit from 3.9 s to 4.2 s (2.1...
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I don't know how to answer this physics question, basically there is a graph, velocity against time graph.

it increases rapidly until 0.7 seconds (2.1 m/s)and starts to level off after that and is a straight line (2.5m/s) after 1 second. It sarts to decrease by levelling off a bit from 3.9 s to 4.2 s (2.1 m/s). And decreases in a steep line rapidly as well to 4.7 seconds to 0meterpersecond.

My question here is what are the answers for these question and why is that ( so please a bit more explanation why certain figures are used would help)

(a) Determine

(i) the initial uniform acceleration of the mass

(ii) the distance travelled by the mass while it is accelerating upwards

the answers are supposed to be:

(a) (i) gradient = = 3.0 ms–2 (1)

(ii) distance is area under graph (to t = 0.1 s)

or × 0.7 × 2.1 0.3 (1) = 1.4(2) m (1)

but why is that, because I don't get the same solution with my calculater.

Thanks for the answers

it increases rapidly until 0.7 seconds (2.1 m/s)and starts to level off after that and is a straight line (2.5m/s) after 1 second. It sarts to decrease by levelling off a bit from 3.9 s to 4.2 s (2.1 m/s). And decreases in a steep line rapidly as well to 4.7 seconds to 0meterpersecond.

My question here is what are the answers for these question and why is that ( so please a bit more explanation why certain figures are used would help)

(a) Determine

(i) the initial uniform acceleration of the mass

(ii) the distance travelled by the mass while it is accelerating upwards

the answers are supposed to be:

(a) (i) gradient = = 3.0 ms–2 (1)

(ii) distance is area under graph (to t = 0.1 s)

or × 0.7 × 2.1 0.3 (1) = 1.4(2) m (1)

but why is that, because I don't get the same solution with my calculater.

Thanks for the answers

Update:
sorry I meant the asnwers are:

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

Update 2:
sorry I meant the asnwers are:

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

Update 3:
sorry I meant the asnwers are:

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

Update 4:
sorry I meant the asnwers are:

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

i) gradient: 2.1m/s/0.7sec = 3ms

ii) 0.7 x 2.1 x ((2.1+2.5)/2) 0.3 = 1.4(2) m (1)

Update 5:
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1...
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guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1 when the graph starts at 0 ? Also how will I ever be sure that they want the whole distance of the graph,is it because it is mentioned that that this graph shows only how it's lift up so the whole graph indicates of acceleration upwards? So when will I know when the fraph is decelerating ? can't I use another method by just calculating the are underneath the graph, whu doesn't this method work?

Thanks again for the answe

Thanks again for the answe

Update 6:
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1...
show more
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1 when the graph starts at 0 ? Also how will I ever be sure that they want the whole distance of the graph,is it because it is mentioned that that this graph shows only how it's lift up so the whole graph indicates of acceleration upwards? So when will I know when the fraph is decelerating ? can't I use another method by just calculating the are underneath the graph, whu doesn't this method work?

Thanks again for the answe

Thanks again for the answe

Update 7:
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1...
show more
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1 when the graph starts at 0 ? Also how will I ever be sure that they want the whole distance of the graph,is it because it is mentioned that that this graph shows only how it's lift up so the whole graph indicates of acceleration upwards? So when will I know when the fraph is decelerating ? can't I use another method by just calculating the are underneath the graph, whu doesn't this method work?

Thanks again for the answe

Thanks again for the answe

Update 8:
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1...
show more
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1 when the graph starts at 0 ? Also how will I ever be sure that they want the whole distance of the graph,is it because it is mentioned that that this graph shows only how it's lift up so the whole graph indicates of acceleration upwards? So when will I know when the fraph is decelerating ? can't I use another method by just calculating the are underneath the graph, whu doesn't this method work?

Thanks again for the answe

Thanks again for the answe

Update 9:
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1...
show more
guys, Im sorry , I left out something again: A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time., that's the beginning of it. Thanks Dancing imu, but in the anwer it sates that 2 in bracket so it's 1.4(2), not 1.43. And how is U=2.1 when the graph starts at 0 ? Also how will I ever be sure that they want the whole distance of the graph,is it because it is mentioned that that this graph shows only how it's lift up so the whole graph indicates of acceleration upwards? So when will I know when the fraph is decelerating ? can't I use another method by just calculating the are underneath the graph, whu doesn't this method work?

Thanks again for the answe

Thanks again for the answe

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