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A car is moving with uniform acceleratio...

A car is moving with uniform acceleration along a straight line between two stops X and Y. Its speed at X and Y are `2 ms^-1 and 14 ms^-1` , Then

A

its speed at mid-point of XY is `10 ms^-1`

B

its speed at a point A such that `XA : AY = 1 :3` is `5 ms^-1`

C

the time to go from X to the mid-point of XY is double of that to go from mid-point to Y

D

the distance travelled in first half of the total time is half of the distance travelled in the second half of the time

Text Solution

Verified by Experts

The correct Answer is:
A, C
`(14)^2=(2)^2+2 as`
`:. 2as=192` units
At mid point, `v^2=(2)^2+2a(s/2)`
`=4+192/2=100`
`:. v=10 m//s`
`XA:AY=1:3`
`:. XA=1/4s` and `AY=3/4s`
`v_1^2=(2)^2+2a(s/4)`
`=4+192/4=52`
`:. v_1=sqrt52!=5 m//s`
`10=2+a t_1 (v=u+at)`
`:. t_1=8/a 14=10+a t_2`
`:. t_2=4/a` or `t_1=2t_2`
`S_1=(2t)+1/2a(t^2)`=distance travelled in first half
`S_2=2(2t)+1/2a(2t)^2`
`S_3=S_2-S_1`=distance travelled in second half
We can see that, `S_1!=(S_3)/2`
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