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2.Motion in Straight Line
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A particle has initial velocity $10\,\, m/s$. It moves due to constant retarding force along the line of velocity which produces a retardation of $5\,\, m/s^2$. Then
Athe maximum displacement in the direction of initial velocity is $10 \,\,m$
Bthe distance travelled in first $3$ seconds is $7.5 \,\,m$
Cthe distance travelled in first $3$ seconds is $12.5\,\, m$
DBoth $(A)$ and $(C)$
Solution
if body has velocity and acceleration in opposite direction then initial body will move in direction if velocity with decreasing speed and then come to rest and will start moving in direction of acceleration with increasing speed.
so initial velocity of body is $10 \mathrm{m} / \mathrm{s}$ and acceleration of the object is $-5 \mathrm{m} / \mathrm{s}^{2}$
so time taken to stop once is 2 sec. using $v=u+a t$ displacement at that time is $10 \mathrm{m}$. using $S=u t+\frac{1}{2} a t^{2}$ or by $v^{2}-u^{2}=2 a s$
after this the body will start moving in reverse direction so maximum displacement is $10 \mathrm{m}$
and distance covered in next $1\, sec$ id calculate by $S=u t+\frac{1}{2} a t^{2}$
so $D_{2}=2.5 m$ so total distance covered in first 3 sec is $10+2.5=12.5$
so best possible answers are $A,C.$
so initial velocity of body is $10 \mathrm{m} / \mathrm{s}$ and acceleration of the object is $-5 \mathrm{m} / \mathrm{s}^{2}$
so time taken to stop once is 2 sec. using $v=u+a t$ displacement at that time is $10 \mathrm{m}$. using $S=u t+\frac{1}{2} a t^{2}$ or by $v^{2}-u^{2}=2 a s$
after this the body will start moving in reverse direction so maximum displacement is $10 \mathrm{m}$
and distance covered in next $1\, sec$ id calculate by $S=u t+\frac{1}{2} a t^{2}$
so $D_{2}=2.5 m$ so total distance covered in first 3 sec is $10+2.5=12.5$
so best possible answers are $A,C.$
Standard 11
Physics
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