1. q6 b) iii) of this paper [ http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_2.pdf]
2: q7c of the same paper.
3. q11 of the same aper
2: q7c of the same paper.
3. q11 of the same aper
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Maths Helpp
- Thread starter Sehrish
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Q.6)b)iii) Upper bound of mass of salt in 100g of beans will be 1.0 + 0.05= 1.05g
Upper bound of mass of salt in 210g of beans will therefore be 1.05/100 x 210= 2.205g
hence the Upper bond of the mass of salt in the tin is 2.205g
Q.7)c) To find the number of litres of fuel in the cylinder, you find the are of segment DBC and multiply it by the length of the cylinder. Now you already found the length in part a and the area in part b.
so Volume of Fuel=0.498x9.82=4.890.
so the number of litres wil be= 4.890x1000= 4890litres ( one cubic metre is 1000 litre)
Q.11) Distance travelled by the object B for the first 5 seconds is the are under the graph of object b for the first 5 seconds. So distance travelled= 1/2 x 5 x 20= 50m
ii) speed is calculated by distance/time. So distance travelled in the first 10seconds= 50+ 5x20=150m. average speed= 150/10= 15m/s
iii) where the two speed time graphs intersect is the point where A and B have the same speed.for this we use the concept of similar triangles. Let t be the time taken at which B has the same speed as A.
t/5=12/20. so t=3 seconds. Therefore both the objects have the same speed as t=3seconds
iv) Let t be the time at which Both A and B have travelled the same distance. So the distance travelled by A will be 12xt (applying the formula distance=speed xtime)= 12t
distance travelled by object B= 1/2 x 20 x(t+(t-5))= 20t -50
both the objects have travelled the same distance so, 12t=20t-50. hence we calculate the time to be 6.25 seconds.
b) d1 is the distance travelled by object B in 5 seconds which is 50m. D2 is the distance travelled by the Object B in 10 seconds which is 150m.
ii) it represents the constant speed of the object B from 5seconds to 10seconds.
iii) When t=2.5seconds, speed of B is 10m/s. So the gradient is 10m/s
c) Deceleration of object A= 12-0/9= 4/3m/s^2
If B comes to rest at t seconds and as both the objects slowed down at the same constant rate, 4/3=20-0/t-10. So t=25seconds. Hence B comes to rest at t=25seconds.
I hope you understand now
Upper bound of mass of salt in 210g of beans will therefore be 1.05/100 x 210= 2.205g
hence the Upper bond of the mass of salt in the tin is 2.205g
Q.7)c) To find the number of litres of fuel in the cylinder, you find the are of segment DBC and multiply it by the length of the cylinder. Now you already found the length in part a and the area in part b.
so Volume of Fuel=0.498x9.82=4.890.
so the number of litres wil be= 4.890x1000= 4890litres ( one cubic metre is 1000 litre)
Q.11) Distance travelled by the object B for the first 5 seconds is the are under the graph of object b for the first 5 seconds. So distance travelled= 1/2 x 5 x 20= 50m
ii) speed is calculated by distance/time. So distance travelled in the first 10seconds= 50+ 5x20=150m. average speed= 150/10= 15m/s
iii) where the two speed time graphs intersect is the point where A and B have the same speed.for this we use the concept of similar triangles. Let t be the time taken at which B has the same speed as A.
t/5=12/20. so t=3 seconds. Therefore both the objects have the same speed as t=3seconds
iv) Let t be the time at which Both A and B have travelled the same distance. So the distance travelled by A will be 12xt (applying the formula distance=speed xtime)= 12t
distance travelled by object B= 1/2 x 20 x(t+(t-5))= 20t -50
both the objects have travelled the same distance so, 12t=20t-50. hence we calculate the time to be 6.25 seconds.
b) d1 is the distance travelled by object B in 5 seconds which is 50m. D2 is the distance travelled by the Object B in 10 seconds which is 150m.
ii) it represents the constant speed of the object B from 5seconds to 10seconds.
iii) When t=2.5seconds, speed of B is 10m/s. So the gradient is 10m/s
c) Deceleration of object A= 12-0/9= 4/3m/s^2
If B comes to rest at t seconds and as both the objects slowed down at the same constant rate, 4/3=20-0/t-10. So t=25seconds. Hence B comes to rest at t=25seconds.
I hope you understand now