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Physics Problems help please

Hi to anyone reading this. Please help me with the questions below. I'm finding physics quite difficult and I have an exam in a week!
A cyclist(A) traveling at a speed of 8.0 ms-1 passes another cyclist (B) who is at rest. It takes B 3.0 seconds to get started just as A passes her and then she can accelerate at 1.0 m s-2 for 10 s. After this time, she travels with a constant velocity. Calculate

i.The cyclist B’s final speed.
1.0 ms-2 x 10 s = 10 ms-1

ii.The distance between the cyclists when B starts.
8.0 ms-1 x 3 s = 24 m

iii.How long does it take for B to catch up with A.

iv.How far have they traveled when B catches up with A.
Original post by elef95
Hi to anyone reading this. Please help me with the questions below. I'm finding physics quite difficult and I have an exam in a week!
A cyclist(A) traveling at a speed of 8.0 ms-1 passes another cyclist (B) who is at rest. It takes B 3.0 seconds to get started just as A passes her and then she can accelerate at 1.0 m s-2 for 10 s. After this time, she travels with a constant velocity. Calculate

i.The cyclist B’s final speed.
1.0 ms-2 x 10 s = 10 ms-1

ii.The distance between the cyclists when B starts.
8.0 ms-1 x 3 s = 24 m

iii.How long does it take for B to catch up with A.

iv.How far have they traveled when B catches up with A.


let t=0 be when B starts accelerating. So the distance that A travels is given by 24 + 8t. Now we do the same for B. we need to calculate the distance travelled by B when accelerating. t=10 u=0 a=1 s=ut + 0.5at^2 so s= 0x10 +0.5x1x10 . s=50 After 10 seconds B travels at a constant 10ms^-1 so the distance travelled by B is given by the equation 50+10(t-10) simplified to 10t-50. B overtakes A when they travel the same distance so 10t-50 =24 +8t solve for t to get t=37.

For iv just sub in the value of t into the equations we found in iii. So for A 24+8x37. distance travelled =320. Verify using equation for B. 10x37 -50 =320

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