Home › NEET UG › Physics › Motion in a Straight Line › A car decelerates from $20$ m/s to rest in $50$ …
A car decelerates from $20$ m/s to rest in $50$ m. Its acceleration magnitude is:
A$2$ m/s$^2$, taking the stopping time to be exactly 10 s
B$8$ m/s$^2$, twice the simple time-based estimate value
C$4$ m/s$^2$, since $v^2 = u^2 + 2as$ gives $a = 4$
D$10$ m/s$^2$, equal to gravitational acceleration $g$
Answer & Solution
Correct answer: C. $4$ m/s$^2$, since $v^2 = u^2 + 2as$ gives $a = 4$
$0 = 20^2 - 2a\cdot 50 \Rightarrow a = 4$ m/s$^2$.
Related questions
A ball is thrown up with initial velocity $u$. It reaches its highest point at time:For a ball thrown vertically upwards and returning to the ground, the velocity-time graph:Displacement differs from distance in that displacement:For uniform motion, velocity is represented by the gradient of a:Which quantities need a sign convention fixed before they are assigned values?In uniform motion, how does instantaneous velocity vary with time?Why can average speed exceed the magnitude of average velocity?Speed differs from velocity in that speed does not carry: