A particle has an initial velocity of 10 , m/ | vedclass

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(D) Given: Initial velocity $u = 10 \, m/s$,acceleration $a = -5 \, m/s^2$. First,we find the time $t_0$ taken to come to rest: $v = u + at \implies 0

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

(D) Given: Initial velocity $u = 10 \, m/s$,acceleration $a = -5 \, m/s^2$. First,we find the time $t_0$ taken to come to rest: $v = u + at \implies 0 = 10 - 5t_0 \implies t_0 = 2 \, s$. The maximum displacement $S_{max}$ in the direction of initial velocity is calculated using $v^2 - u^2 = 2aS$: $0^2 - 10^2 = 2(-5)S_{max} \implies S_{max} = 10 \, m$. Thus,option $(A)$ is correct. To find the distance travelled in $3 \, s$: In the first $2 \, s$,the distance is $10 \, m$. In the remaining $1 \, s$ (from $t=2$ to $t=3$),the particle starts from rest $(u'=0)$ and moves with acceleration $a=5 \, m/s^2$: $S' = u't + \frac{1}{2}at^2 = 0 + \frac{1}{2}(5)(1)^2 = 2.5 \, m$. Total distance in $3 \, s = 10 + 2.5 = 12.5 \, m$. Thus,option $(C)$ is correct. Since both $(A)$ and $(C)$ are correct,the final answer is $(D)$.

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

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