Help with motions/suvat question ?

Hi I'm stuck on this question I have answered 2 parts of it but not the rest. I'm sure im making a basic error but I can't figure it out, please help Thanks :smile:

Question:

A cyclist accelerated from rest at a constant acceleration of 0.4m/s^2 for 20 seconds, then stopped pedaling and slowed to a standstill at constant deceleration over a distance of 260m.

a) calculate (i) the distance travelled by the cyclist in the first 20s (answer is 80m) and (ii) the speed of the cyclist at the end of this time (answer is 8.0 ms^-1)

B) Calculate (i) the time taken to cover the distance of 260m after she stopped pedalling,

(ii) her deceleration during this time

I can't figure out part B (i) and (ii) please help, Thank you so much :smile:

Reply 1

Dinasaurus

22

Sorry if this is wrong, but could you do s=((u+v)/2)t

So 260=4t therefore t=65s?

Not sure if that is sensible, or not.

Then deceleration would just be v=u+at

So 0=8+65a, so -8=65a. Therefore -8/65=a

Sorry if it's wrong, just a thought.

Reply 2

uberteknik

21

Original post by Dinasaurus

Sorry if this is wrong, but could you do s=((u+v)/2)t

So 260=4t therefore t=65s?

Not sure if that is sensible, or not.

Then deceleration would just be v=u+at

So 0=8+65a, so -8=65a. Therefore -8/65=a

Sorry if it's wrong, just a thought.

Yes, that works.

Can also do:

v^2 = u^2 + 2as

rearranging for acceleration:

a = \frac{v^2 - u^2}{2s}

and substituting values:

a = \frac{0 - 8^2}{2 \rm x 260} = -0.123 ms^{-2}

then use

v = u + at

t = \frac{v - u}{a} = \frac{0 - 8}{-0.123} = 65

seconds

Reply 3

Username002

2

After calculating time use S=ut+1/2at^2
You have s the distance taken to stop = 260m, u from the final velocity of the first part of the journey = 8ms^-1 and you have t = 65s, rearrange for a.

Reply 4

delta.charlie

2

In questions such as this it always very helpful to write down what you know. Like:
u=8 m/s
v=0, since the cyclist comes to rest
s=260m
t=?
Now u can find the time using s=((u+v)/2)t and deceleration by using v=u+at as mentioned by Dinasaurus.
I also got the same answers mentioned by other users

Reply 5

Cinna21

OP

12

Original post by delta.charlie

In questions such as this it always very helpful to write down what you know. Like:
u=8 m/s
v=0, since the cyclist comes to rest
s=260m
t=?
Now u can find the time using s=((u+v)/2)t and deceleration by using v=u+at as mentioned by Dinasaurus.
I also got the same answers mentioned by other users

Thank you so much everyone I really appreciate the help and yes I was making a silly mistake aha, but Thank you so much I can now live in peace :P Thanks :smile:

Help with motions/suvat question ?

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