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Kinematics help

Screenshot (3).png
Im really confused about this kinematics question?
For the "certain 4 seconds", use s = vt - 1/2 at^2
Then for the next 5 seconds, use s = ut + 1/2 at^2, but remember that your initial velocity at the start of the 5 seconds is the same as the final velocity at the end of the 4 seconds, so you can let u = v for the second equation. Rearrange and solve as simultaneous equations.
Original post by Lion2

Im really confused about this kinematics question?


If it starts from rest, then the velocity at time tt is given by v=atv=at and the displacement is given by s=12at2s=\frac{1}{2}at^2.

At some time t=Tt=T the particle has displacement from the origin as s1=12aT2s_1=\frac{1}{2}aT^2, and 4 seconds later it has a displacement s2=12a(T+4)2s_2 = \frac{1}{2}a(T+4)^2.

We know that the difference between the two displacements is 12m therefore s2s1=12=12a[(T+4)2T2]s_2 - s_1 = 12 = \frac{1}{2}a[(T+4)^2-T^2].

Doing the same with the next piece of information yields 30=12a[(T+9)2(T+4)2]30 = \frac{1}{2}a[(T+9)^2-(T+4)^2]

You can use the two to determine TT, and hence aa. Hence answer the rest of the questions.
Original post by RDKGames
If it starts from rest, then the velocity at time tt is given by v=atv=at and the displacement is given by s=12at2s=\frac{1}{2}at^2.

At some time TT the particle has displacement from the origin as s1=12aT2s_1=\frac{1}{2}aT^2, and 4 seconds later it has a displacement s2=12a(T+4)2s_2 = \frac{1}{2}a(T+4)^2.

We know that the difference between the two displacements is 12m therefore s2s1=12=12a[(T+4)2T2]s_2 - s_1 = 12 = \frac{1}{2}a[(T+4)^2-T^2].

Doing the same with the next piece of information yields 30=12a[(T+9)2(T+4)2]30 = \frac{1}{2}a[(T+9)^2-(T+4)^2]

You can use the two to determine TT, and hence aa. Hence answer the rest of the questions.


Would my method still work?
Original post by 3pointonefour
Would my method still work?


It's hard to see what you're trying to do.

So have a go and see if you get the right answers.
Original post by RDKGames
It's hard to see what you're trying to do.

So have a go and see if you get the right answers.


Well, using both methods I came to the same value for a. I guess there's more than one way to solve a problem :biggrin:
Reply 6
Original post by RDKGames
If it starts from rest, then the velocity at time tt is given by v=atv=at and the displacement is given by s=12at2s=\frac{1}{2}at^2.

At some time t=Tt=T the particle has displacement from the origin as s1=12aT2s_1=\frac{1}{2}aT^2, and 4 seconds later it has a displacement s2=12a(T+4)2s_2 = \frac{1}{2}a(T+4)^2.

We know that the difference between the two displacements is 12m therefore s2s1=12=12a[(T+4)2T2]s_2 - s_1 = 12 = \frac{1}{2}a[(T+4)^2-T^2].

Doing the same with the next piece of information yields 30=12a[(T+9)2(T+4)2]30 = \frac{1}{2}a[(T+9)^2-(T+4)^2]

You can use the two to determine TT, and hence aa. Hence answer the rest of the questions.

Thank you so much! I got the answers in the end using your method
Original post by Lion2
Screenshot (3).png
Im really confused about this kinematics question?

The answers are:
0.667m/s^2
1.67m/s
2.08m

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