Question

A particle has initial velocity $10m/s$ . It moves due to constant retarding force along the line of velocity which produces a retardation of $5m/s^2$. Then

• A. the maximum displacement in the direction of initial velocity is $10m$
  
• B. the distance travelled in first $3$ seconds is $7.5m$
  
• C. the distance travelled in first $3$ seconds is $12.5m$
  
• D. the distance travelled in first $3$ seconds is $17.5m$
  

Answer 1

Answer: (A), (C)

The correct options are
A the maximum displacement in the direction of initial velocity is $10m$
C the distance travelled in first $3$ seconds is $12.5m$

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 $10m/s$ and acceleration of the object is $-5m/s^2$

so time taken to stop once is 2 sec. using $v=u+at$

displacement at that time is 10m. using $S=ut+\frac{1}{2}a{t}^2$ or by $v^2-u^2=2as$

after this the body will start moving in reverse direction so maximum displacement is 10m.

and distance covered in next 1 sec id calculate by $S=ut+\frac{1}{2}a{t}^2$

so $D_2=2.5m$ so total distance covered in first 3 sec is $10+2.5=12.5$

so best possible answers are A,C.