MJLover's A level physics thread

For my brother, he's studying his A Levels -

A cyclist is travelling at a constant speed of 3 ms-1 as he starts to roll down a hill. He rolls down the hill with a constant acceleration. During the third second, he travels at a distance of 6 m.

a) Calculate the cyclist's acceleration.
b) Calculate how far he travels during the 4th second.

I would appreciate explanations of how you understood and got the answer

(edited 7 years ago)

bumpity

acceleration = v-u/t

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Reply 3

AdeptDz

Use suvat equations
S = displacement / distance
U = initial velocity
V = final velocity
A = acceleration
T = time

V = u+at

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Reply 4

MJlover

Original post by AdeptDz

acceleration = v-u/t

Posted from TSR Mobile

Can I please ask if you could directly give a solution to the answer?

Reply 5

AdeptDz

I got a = 0, But im not sure i jsut started learning this too.
I assumed final velocity was 0 as it doesn;t explicitly say it but "constant speed of 3ms-1" means he didnt slow down so im guessing he was going the same speed all the way up to 3 seconds so same velocity, but ill do question b now
a=v-u/t
a=3-3/3 = 0

I got b is 9 or 12, realistically i got 12, but i probably made a mistake somewhere so hopefully someone comes and clears this up

(edited 7 years ago)

What did you get?

Basically for a) all you do is use the suvat equations and solve a simultaneous equation using the distance at 2 seconds (s) and the distance at 3 seconds (s+6):
u = 3ms^-1
a = constant
t = 2/3 seconds
s = distance (varied)

Equation used: s = ut + 0.5(at²)
Simultaneous equations:
s = 3(2) + 0.5a(2)²
s+6 = 3(3) + 0.5a(3)²

When I solved it I got the acceleration as 1.2ms^-2

Once a) was done b) is simple as the acceleration is known.
Distance for the 4th second (t=3 to t=4)
when t=3,
s = ut + 0.5(at²)
s = 3(3) + 0.5(1.2)(3)²= 14.4m
when t=4,
s = 3(4) + 0.5(1.2)(4)²= 21.6m

So the distance travelled in the 4th second is 21.6-14.4 = 7.2m :smile: