------------*****Help with math M1*****------------

[XPFMember]

Assalamoalaikum!!

Can sumbody plz help me with Q:2 i of http://www.xtremepapers.me/CIE/Internat ... 2_qp_4.pdf

Jazak Allah Khair

 

[aliya_zad]

Walaikumusalam.

In part i there are 2 different kinds of motion involved. For the first 20s its constant motion and for the next 4 its decelerating.
For the first part apply speed=distance/time and u get the distance as 140m.
For the second part uv gotta use equations of motion, where u=7, v=0 and t=4.
s= 1/2(u+v) x t and u get s as 14m.
total s=154.

ii) here uv gotta find the ()distand travelled for the first 20s which uv alredy found+ 20m)- distance travelled between 30-40s(found using area under graph).
coz its travelling in opp direction.
hope u understand

 

[XPFMember]

no i mean how do we prove dist. is greater than 154...i mean do we write some sentence or something..??
Jazak Allah Khair..for the quick response

 

[anahita16]

Total distance moved if deceleration were uniform
= Area under trapezium from t=o to t=24
= 0.5*(20+24)*7
= 154m
But since deceleration was not uniform, the actual area under graph is greater than the area under the trapezium.
Therefore, the total distance run by the man is greater than 154m.
Hence shown

 

[anahita16]

[........
For the second part uv gotta use equations of motion, where u=7, v=0 and t=4.
s= 1/2(u+v) x t and u get s as 14m.
total s=154.
...
The equations of motion cannot be ued for the second part because the deceleration is not constant, that is, it changes between t=20s and t=24s. The equations of motion are valid only in cases of constant acceleration.