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Please help with as level physics question.....

An object is dropped from a cliff. How far does the object fall in the third second? Assume that g=10ms^-2

A) 10 metres
B) 20 metres
C) 25 metres
D) 45 metres

Please help, I have been stuck on this question for days!! Any help would be much appreciated.
The answer I keep getting is 45 metres but the mark scheme says something else :frown:
Reply 1
Avatar for JN17
JN17
10
Well if after every second, it is travelling 10ms^-1 faster than the second before, and that the third second refers to second two -> second three (as in t=2 -> t=3)
at t=2 it is travelling at 20ms^-1, and at t=3 it is travelling at 30ms^-1, from that we can find an average velocity over the third second and hence the distance travelled.
Reply 2
Avatar for Pedals
Pedals
2
I got 25 metres.
We know the variables u (initial velocity) = 0ms^-1 and acceleration = 10ms^-2.
To find the distance travelled in the 3rd second you need to subtract the distance travelled in the first 2 seconds from the distance travelled in 3 seconds.

Using the SUVAT equation S = ut + (1/2)at^2, where u = 0 ms^-1:

distance travelled in the 3rd second = (1/2)*10*3^2 - (1/2)*10*2^2 = 25m.

Hope this helps :smile:
(If you see any mistakes please let me know :smile: )
(edited 7 years ago)
Reply 3
Avatar for sue99
sue99
10
Original post by Jas1947
An object is dropped from a cliff. How far does the object fall in the third second? Assume that g=10ms^-2

A) 10 metres
B) 20 metres
C) 25 metres
D) 45 metres

Please help, I have been stuck on this question for days!! Any help would be much appreciated.
The answer I keep getting is 45 metres but the mark scheme says something else :frown:


Just to check, does the mark scheme say that 25m is the answer?
Too late, someone got there before me :smile: sorry, page was taking ages to load, but if you still need an explanation, don't hesitate to ask :-)
(edited 7 years ago)
Reply 4
Avatar for Minnie16
Minnie16
OP
8
Original post by Pedals
I got 25 metres.
We know the variables u (initial velocity) = 0ms^-1 and acceleration = 10ms^-2.
To find the distance travelled in the 3rd second you need to subtract the distance travelled in the first 2 seconds from the distance travelled in 3 seconds.

Using the SUVAT equation S = ut + (1/2)at^2, where u = 0 ms^-1:

distance travelled in the 3rd second = (1/2)*10*3^2 - (1/2)*10*2^2 = 25m.

Hope this helps :smile:
(If you see any mistakes please let me know :smile: )


Thank you so much! Oh it seems so simple now, i feel stupid being stuck on this for days , thank you so much !
Reply 5
Avatar for Minnie16
Minnie16
OP
8
Original post by sue99
Just to check, does the mark scheme say that 25m is the answer?
Too late, someone got there before me :smile: sorry, page was taking ages to load, but if you still need an explanation, don't hesitate to ask :-)


Yes the mark scheme says 25 metres

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