Question

A car is moving on a straight road with
uniform velocity 10 m/s. Driver applies brake
which produces retardation of 2 m/s2 in how
much time truck will stops.​

Answer 1

$\sf{\blue{\underline{Given:}}}$

Initial velocity,u = 10m/s

Final velocity,v = 0

acceleration = -2m/s^2 (as the car is retarding,the acceleration will be negative)

$\sf{\pink{\underline{To\;Find:}}}$

☆Time taken to stop

$\sf{\red{\underline{Solution:}}}$

Using first equation of motion :

$\boxed{\green{v\;=\;u\;+\;at}}$

where,

v is final velocity

u is initial velocity

a is the acceleration produced

t is time taken

by putting values in above equation,we get :

$\sf{\implies 0\;=10+(-2)* t}$

$\sf{\implies 0\;=\;10-2t}$

$\sf{\implies -10\;=\;-2t}$

$\sf{\implies t = 5 sec }$

hence,the time taken to stop is

$\boxed{\red{t\;=\;5\;sec}}$

Answer 2

Answer:

plz check the attachment above

hope it helps u