Question

A particle which is moving in a straight line with constant acceleration covers distances of 10m and 15m in two successive seconds. Find the acceleration

Answer 1

@TuringTest @UnkleRhaukus @precal

Answer 2

i ve just started learning this at my class...but i know those:)

Answer 3

derivative and calculus correct?

Answer 4

I'll assume you are familiar with the equations of motion, particularly this one
\[s = ut + \frac{1}{2}at^{2}\] Let us take the starting time to be t. In time t -> t+1, it covers 10m and from time t+1 -> t+2 it covers 15m

Now instead of using t+1 -> t+2, i'll make use of the fact that it covered 25m from t->t+2
This is used, because in both cases the initial velocity 'u' is the same.

10 = u(t+1) + 0.5a(t+1)^2 - [ut + 0.5at^2]
25 = u(t+2) + 0.5a(t+2)^2 - [ut + 0.5at^2]

Can you take it from here? The answer is still a couple steps away

Answer 5

listen to FoolAroundMath

Answer 6

define two successive seconds?

Answer 7

t->t+1 is the first second, t+1->t+2 is the second successive second

Answer 8

\[\vec a=\frac{\vec v_1+\vec v_2}{2}=\frac{\frac{\Delta x_1}{\Delta t_1}+\frac{\Delta x_2}{\Delta t_2}}{2}=\frac{\frac{10\text m}{1\text s}+\frac{15\text m}{1\text s}}{2}\]

Answer 9

@FoolAroundMath why did u take 0.5?

Answer 10

Isn't that the second equation of motion?
\[s = ut + \frac{1}{2}at^{2}\]
adding a bit more to my answer,
The displacement in t->t+1 = total displacement from time 0 to t+1 - total displacement from time 0 to t.

The displacement in t->t+2 = total displacement from time 0 to t+2 - total displacement from time 0 to t

Answer 11

isnt there any other easy way to do this question?

Answer 12

Can't think of anything simpler at the moment. :(

Answer 13

hm anyway thanks :)

Answer 14

@ash2326 @TuringTest

Answer 15

no...><

Answer 16

wait a minute I m seeing formula again:)

Answer 17

yeah i was right
\[\overrightarrow a=\frac{\overrightarrow S_2 -\overrightarrow S_1}{t^2}\]

Answer 18

\[\overrightarrow a= \frac{15-10}{1^2}=5 m/s^2\]

Answer 19

@thushananth01 gt it?

Answer 20

but how, i never learnt tht equation?

Answer 21

wait a minute

Answer 22

\[\overrightarrow a=\frac{\overrightarrow v_2-\overrightarrow v_1}{t}\]\[\implies \overrightarrow a = \frac{\frac {\overrightarrow S_2}{t}-\frac{\overrightarrow S_1}{t}}{t}\]\[\implies \overrightarrow a = \frac{\overrightarrow S_2-\overrightarrow S_1}{t^2}\]

Answer 23

gt it now?

Answer 24

yes thnks :)

Answer 25

yw:)
so correct answer is 5m/s^2

Answer 26

yes

Answer 27

^_^

Answer 28

thats what i was trying to do

Answer 29

@ganeshie8

Answer 30

mahesh .. so .. why is the time 1 ... it should be 2