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M1 help (again) - sorry
Sorry guys, I'm stuck again.
Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9ms^-1 and A is overtaking B at a speed of 10ms^-1. At the moment when they are level they see a traffic light turn red 108m ahead. A cycles for T seconds and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the traffic light at the same instant.
Calculate the time during which B was decelerating.
Calculate the value of T.
I really don't get this.
Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9ms^-1 and A is overtaking B at a speed of 10ms^-1. At the moment when they are level they see a traffic light turn red 108m ahead. A cycles for T seconds and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the traffic light at the same instant.
Calculate the time during which B was decelerating.
Calculate the value of T.
I really don't get this.
please help
Widowmaker
Sorry guys, I'm stuck again.
Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9ms^-1 and A is overtaking B at a speed of 10ms^-1. At the moment when they are level they see a traffic light turn red 108m ahead. A cycles for T seconds and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the traffic light at the same instant.
Calculate the time during which B was decelerating.
Calculate the value of T.
I really don't get this.
Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9ms^-1 and A is overtaking B at a speed of 10ms^-1. At the moment when they are level they see a traffic light turn red 108m ahead. A cycles for T seconds and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the traffic light at the same instant.
Calculate the time during which B was decelerating.
Calculate the value of T.
I really don't get this.
For the first bit you work out how far B goes in the 6 seconds so that's 9 X 6 = 54. The total distance it goes is 108m, therefore when it is decelerating it travels 108m - 54m = 54 m.
Using v^2 = u^2 + 2as
0^2 = 9^2 + 2 x a x 54
0= 81 + 108a
so a = - 81/108
a = - 0.75 m/s
using v = at + u
0 = -0.75t + 9
t = -9/ -0.75
t= 12 seconds
I think that's right... I'm working out the second bit now...
Widowmaker
Sorry guys, I'm stuck again.
Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9ms^-1 and A is overtaking B at a speed of 10ms^-1. At the moment when they are level they see a traffic light turn red 108m ahead. A cycles for T seconds and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the traffic light at the same instant.
Calculate the time during which B was decelerating.
Calculate the value of T.
I really don't get this.
Two cyclists A and B are travelling in the same direction along a straight road. B is travelling at a speed of 9ms^-1 and A is overtaking B at a speed of 10ms^-1. At the moment when they are level they see a traffic light turn red 108m ahead. A cycles for T seconds and then decelerates uniformly; B cycles for 6s and then decelerates uniformly. A and B stop at the traffic light at the same instant.
Calculate the time during which B was decelerating.
Calculate the value of T.
I really don't get this.
B: in 6s travels 9*6 = 54m
therefore 54m left
v^2 = u^2 + 2as
0= 81 +2a(54)
108a = -81
a= -0.75
v=u+at
0=9-0.75t
t=12s
A: s =10T + ut +0.5at^2
t= 12+6 (from part one)
108 = 10T +10(18-T)+0.5a(18-T)^2
v=u+at
0=10+a(18-T)
a=-10/(18-T)
108 = 10T+10(18-T) +0.5(-10/(18-T))(18-T)^2
108 = 10T+180-10T -5(18-T)
5T = 108-180+90 = 18
T = 3.6s
I've probably made a mistake somewhere. Do you know what the answers are?
hitchhiker_13
B: in 6s travels 9*6 = 54m
therefore 54m left
v^2 = u^2 + 2as
0= 81 +2a(54)
108a = -81
a= -0.75
v=u+at
0=9-0.75t
t=12s
A: s =10T + ut +0.5at^2
t= 12+6 (from part one)
108 = 10T +10(18-T)+0.5a(18-T)^2
v=u+at
0=10+a(18-T)
a=-10/(18-T)
108 = 10T+10(18-T) +0.5(-10/(18-T))(18-T)^2
108 = 10T+180-10T -5(18-T)
5T = 108-180+90 = 18
T = 3.6s
I've probably made a mistake somewhere. Do you know what the answers are?
therefore 54m left
v^2 = u^2 + 2as
0= 81 +2a(54)
108a = -81
a= -0.75
v=u+at
0=9-0.75t
t=12s
A: s =10T + ut +0.5at^2
t= 12+6 (from part one)
108 = 10T +10(18-T)+0.5a(18-T)^2
v=u+at
0=10+a(18-T)
a=-10/(18-T)
108 = 10T+10(18-T) +0.5(-10/(18-T))(18-T)^2
108 = 10T+180-10T -5(18-T)
5T = 108-180+90 = 18
T = 3.6s
I've probably made a mistake somewhere. Do you know what the answers are?
No, you are right. thanks a lot
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