Question

A car starts from rest and attains a velocity of 10 metre per second in 40 seconds .The driver applies brakes and slows down the car to 5 metre per second in 10 seconds. Find the acceleration of the car in both the cases?.​

Answer 1

initial velocity=0

final velocity = 10 m/s

time=40 s

acceleration = final velocity - initial velocity / time = ( 10-0)/40=10/40=1/4 m/s^2

initial velocity = 10 m/s

final velocity = 5m/s

time=10s

acceleration = final velocity - initial velocity / time = ( 5-10)/10= -5/10= -1/2 m/s^2

hope my answer helps you!

Answer 2

CASE-I

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time taken (t) = 40 s

From first equation of motion we get:

$\bf \leadsto v = u + at \\ \\ \rm \leadsto 10 = 0 + 40a \\ \\ \rm \leadsto 40a = 10 \\ \\ \rm \leadsto a = \dfrac{ \cancel{10}}{4 \cancel{0}} \\ \\ \rm \leadsto a = \dfrac{1}{4} \: m {s}^{ - 2} \\ \\ \rm \leadsto a = 0.25 \: m {s}^{ - 2}$

$\therefore$ Acceleration (a) = 0.25 m/s²

CASE-II

Initial velocity (u) = 10 m/s

Final velocity (v) = 5 m/s

Time taken (t) = 10 s

From first equation of motion we get:

$\bf \leadsto v = u + at \\ \\ \rm \leadsto 5 = 10 + 10a \\ \\ \rm \leadsto 10a = 5 - 10 \\ \\ \rm \leadsto 10a = -5 \\ \\ \rm \leadsto a = -\dfrac{5}{10} \\ \\ \rm \leadsto a =- \dfrac{1}{2} \: m {s}^{ - 2} \\ \\ \rm \leadsto a = -0.5 \: m {s}^{ - 2}$

$\therefore$ Acceleration (a) = -0.5 m/s²

Note: Negative sign in acceleration denotes the retarding motion of the body (Retardation).