Question

A particle accelerated uniformly from rest at 6.0m/s² for 8s,then decelerates uniformly to rest in the next 5s.D
etermine the magnitude of the deceleration.

Answer 1

Explanation:

in first case a=6. u=0 v=? and t=8s

applying first equation of motion

v=u+at

v=0+6×8

v=48

in second case u=48 v=0 and t=5

v=u+at

0=48+a×5

a= -9.6 m/s2

hence magnitude of retardation is 9.6 m/s2

hope you got it

Answer 2

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

Let the acceleration be ${a_1 = 6 m/s^2}$

and, deceleration be ${a_2 = x m/s^2}$

${u_1 = 0 ; v_2 = 0 m/s}$

$\huge{\underline{Explanation:-}}$

Using the first kinematics equation we get,

$\huge{\boxed{\boxed{v_1 = u_1 + a_1t}}}$

Substituting the values

${v_1 = 0 + 6 \times 8}$

${v_1= 48 m/s}$

Now this velocity will be initial velocity and the final velocity becomes zero.

${\therefore v_1 = u_2}$

Again first kinematics equation.

$\huge{\boxed{\boxed{a_2 = \frac{v_2 - u_2}{t}}}}$

${a_2 = \frac{0 - 48}{5}}$

${a_2 = - 9.6 m/s^2}$

Taking its magnitude (-ve) sign will be removed.

$\huge{\boxed{\boxed{a_2 = 9.6 m/s^2}}}$

so, magnitude of deceleration is 9.6 m/s^2.