A particle of mass $m$ moves on the $x$ ‐axis as follows: it starts from rest at $t = 0$ from the point $x = 0,$ and comes to rest at $t = 1$ at the point $x = 1$ . No other information is available about its motion at intermediate times $(0 < t < 1)$ . If $α$ denotes the instantaneous acceleration of the particle, then,

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$α$ cannot remain positive for all $t$ in the interval $0 \leq t \leq 1.$

$| α |$ must be $\geq 4$ at some point or points in its path.

$| α |$ cannot exceed 2 at any point in its path.

$α$ must change sign during the motion, but no other assertion can be made with the information given.

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The positions of the particle at $t = 0$ and at    $t = 1 s$ are
   $x (0) = 0 m,$ $x (1) = 1 m .$
The particle is at rest at    $t = 0$ and comes to rest at $t = 1 s$ . Thus, initial and final velocities of the particle are
$v (0) = 0,$ $v (1) = 0$ . (1)
The velocity $v (1) = 0$ of a particle having instantaneous acceleration $α$ is given by
$v (1) = v (0) +_{0}^{1} α d t$ . (2)
Question Image
The equations (1) and (2) give, $_{0}^{1} α d t = 0$ . Thus, area under the $α - t$ curve is zero. If $α = 0$ for all $0 < t < 1$ then area becomes zero but this also makes $x (1) = 0,$ which is not true. If $α > 0$ for all $0 < t < 1$ then
area under the curve is positive and if $α < 0$ for all    $0 < t < 1$ then area under the curve is negative. Thus, $α$ must change sign at some time in $0 < t < 1$ . Only conclusions that can be made are $_{0}^{1} α d t = 0, α \neq 0,$ and $α$ must change sign (at least once) in $0 < t < 1.$
Question Image
If $α$ change sign only once (say at $t = t_{c}$ ) and $α = α_{1}$ in $0 < t < t_{c}$ and $α = - α_{2}$ in $t_{c} < t < 1$ , where $α_{1}$ and $α_{2}$ are some positive constants, then condition $x (1) = 1 m$ is violated if $α < 4$ at all points. We encourage you to explore area under v‐t graph to show that $α > 4$ at some point(s). Hint: The area under v‐t graph, $\frac{1}{2} \times 1 \times v_{max} = 1$ , gives $v_{max} = 2 m / s .$ Now, $v_{max} = 2 = α_{1} t_{c} = α_{2} (1 - t_{c})$ . If $t_{c} < \frac{1}{2}$ than    $α_{1} > 4$ otherwise $α_{2} > 4.$