The Student Room Group

Mechanics- velocity and acceleration

Iam getting the ans as 63m, but the ans in the text book is 62m. Pls help16068295902981902242141633749707.jpg its the 13th question
(edited 3 years ago)

Scroll to see replies

Reply 1
Not sure about Chris's acceleration. Or the time correction.
(edited 3 years ago)
Reply 2
Original post by mqb2766
Not sure about Chris's acceleration. Or the time correction.


Chris acceleration would be 2m/s^2
Reply 3
Original post by mqb2766
Not sure about Chris's acceleration. Or the time correction.

When t=8, and v=16m/s, a=2m/s^2.
Original post by Shas72
Chris acceleration would be 2m/s^2


You’ve written 1.6 (even though your calculation is using 2 anyways)

Is it definitely 62 in the book?
Reply 5
Original post by GabiAbi84
You’ve written 1.6 (even though your calculation is using 2 anyways)

Is it definitely 62 in the book?

Yeah
Reply 6
Original post by GabiAbi84
You’ve written 1.6 (even though your calculation is using 2 anyways)

Is it definitely 62 in the book?

The distance is 62m and for the second question time is 23.1s
Reply 7
Original post by Shas72
The distance is 62m and for the second question time is 23.1s

Have you got the time correction sorted?
Draw the velocity time graph if necessary.
Reply 8
I don't understand the time for chris. is it (t-5) or (t+5)
Reply 9
if Bradley t=0, Chris starts 5s later than Bradley will be t+5
Original post by Shas72
I don't understand the time for chris. is it (t-5) or (t+5)

You've done this before. Tbh, you just need to see how long he spends at constant speed. It's easy to reason.
Original post by Shas72
The distance is 62m and for the second question time is 23.1s


How can the second part be 23.1s when the first part proved that Chris hadn’t overtaken him by the time he started decelerating at 50s?
Reply 12
I know and you only taught me that it will be t-5.
Reply 13
so I need to first find the time for chris
Original post by Shas72
I know and you only taught me that it will be t-5.

Ive tried to get you to think more, sketch and respond slower.
Reply 15
ok sure. thanks!!
Original post by Shas72
ok sure. thanks!!

Upload your sketch when done.
You don't really need t+/-5. You are given the time intervals.
Reply 17
Original post by mqb2766
Upload your sketch when done.
You don't really need t+/-5. You are given the time intervals.


Ok thanks! Will do that
Reply 18
Original post by mqb2766
Upload your sketch when done.
You don't really need t+/-5. You are given the time intervals.

16068373088993178517004140744081.jpg
16068379783784212000260126767112.jpg
(edited 3 years ago)
Original post by Shas72
16068373088993178517004140744081.jpg
16068379783784212000260126767112.jpg

But they don't start at the same time.

Quick Reply

Latest