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I'm struggling a bit with this question, it seems quite simple but I can't seem to get the right answer.. pls help:
A car accelerates from rest at a rate of 3.5 metres per second squared for 6 seconds; it then travels at a constant speed for a further 8 seconds. Calculate the distance that the car has travelled in this time.
Draw a acceleration distance graph and work out area under it
(edited 5 years ago)
Original post by Kkapoor155
Draw a acceleration distance graph and work out area under it

I've tried but according to the answer book the distance is 231m and I don't get that answer using the graph
Reply 3
Original post by RIP_Katie_2018
I'm struggling a bit with this question, it seems quite simple but I can't seem to get the right answer.. pls help:
A car accelerates from rest at a rate of 3.5 metres per second squared for 6 seconds; it then travels at a constant speed for a further 8 seconds. Calculate the distance that the car has travelled in this time.

Okay so I take it you've done the equations of linear motion as this is in the A Level section, so a=3.5 ms^-2

For the first 6 seconds the distance is given by
S=(at^2)/2

Then for the final 8 seconds the distance is given by
S=ut. T=8 in this

Where u=at t=6 in this


Then sum the distances and hey Presto
(edited 5 years ago)
Original post by M4cc4n4
Okay so I take it you've done the equations of linear motion as this is in the A Level section, so a=3.5 ms^-2

For the first 6 seconds the distance is given by
S=(at^2)/2

Then for the final 8 seconds the distance is given by
S=ut. T=8 in this

Where u=at t=6 in this


Then sum the distances and hey Presto

I got it! Thank you so much :smile:)
Original post by RIP_Katie_2018
I got it! Thank you so much :smile:)


Are you in As-Level ? I need help
Reply 6
Original post by Daniyal Shah
Are you in As-Level ? I need help

What do you need help with?
Original post by M4cc4n4
Okay so I take it you've done the equations of linear motion as this is in the A Level section, so a=3.5 ms^-2

For the first 6 seconds the distance is given by
S=(at^2)/2

Then for the final 8 seconds the distance is given by
S=ut. T=8 in this

Where u=at t=6 in this


Then sum the distances and hey Presto


Well done! you are new here, aren't you? have never faced you in a science forum so far.
Reply 8
Original post by Kallisto
Well done! you are new here, aren't you? have never faced you in a science forum so far.

Not really, I mostly stick to the maths and sometimes physics forums, but have been trying to help more in the physics one recently
Original post by M4cc4n4
What do you need help with?

Can send me Kinematics Notes
Original post by Daniyal Shah
Can send me Kinematics Notes

I already have my notes away sorry!

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