# 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 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

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)

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)

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)

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)

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

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

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

Thanks again for the answe

### 1 Answer

- Dancing ImuLv 67 years agoFavorite Answer
Uniform acceleration equations:

(i) It increases rapidly until 0.7 seconds (2.1 m/s). I assume that means that it accelerates for 0.7 seconds up to a speed of 2.1 m/s. You use the equation

v = u +at

Where:

v = Final velocity

u= Initial velocity

a = Acceleration

t = Time

I have to assume that it started at rest.

So plug in your numbers

v = 2.1 m/s

u = 0.0 m/s

t = 0.7 s

a = unknown

(2.1 m/s) = (0.0 m/s) + [ a * (0.7 s) ]

(2.1 m/s) = [ a * (0.7 s) ]

(2.1 m/s) / (0.7 s) = a

3 m/s^2 = acceleration

****

ii) starts to level off after that and is a straight line (2.5m/s) after 1 second, find distance traveled during its initial acceleration and I assume also its slower acceleration (leveling off)

Use the equation:

s = 0.5 * (u + v) * t

Where

s = Distance

u = Initial velocity

v = Final velocity

t = Time

Given --

s = Unknown

u = 2.1 m/s

v = 2.5 m/s

t = 0.3 s

You get 0.3 seconds because the time different is from the end of the initial acceleration at 0.7 s and a flat line at 1.0 s (1.0 s- 0.7 s= 0.3 s)

So

s = 0.5 * [ (2.1 m/s) + (2.5 m/s) ] * (0.3 s)

s = 0.5 * [ 4.6 m/s ] * (0.3 s)

s1 = 0.69 m

Now, I'm guessing that this is a two-part question and you might want the *total* distance traveled during the entire acceleration. So use this equation:

s = ut + 0.5 * a * t^2

s = Distance

u = Initial velocity

a = Acceleration that you calculated above

t = Time

You have:

s = Unknown

u = 0.0 m/s

t = 0.3

a = 3 m/s^2

s = 0.5 * [ (0.0 m/s) * (0.7 s) ] + [ (3 m/s^2) * (0.7 s)^2 ]

s = 0.5 + [ (3 m/s^2) * (0.49 s^2) ]

s = 0.5 + [ 1.47 m ]

s2 = 0.735 m

So the GRAND TOTAL distance traveled during acceleration

D = s1 + s2

D = (0.69 m) + (0.74 m)

D = 1.43 m

Ta-dah!

Source(s): General science http://hsphysicslab.blogspot.com/2007/04/equations...