Question

A particle starts from rest and had an acceleration of 2m/s2 for 10s. After that it travels for 30s. With constant speed and then undergoes a retardation of 4m/s2 and comes back to rest. The total distance covered by the particle is

Answer 1

U= 0
a=2m/s2
t=10s+30s
=40s
v=4m/s2

v=d/t
4=d/40
4×40=d
d=160

Answer 2

$\boxed{\rm{\pink{Given \longrightarrow }}}$

$\sf\purple{↬}$Initial speed is 0m/s

$\sf\purple{↬}$Acceleration is 2m/s²

$\sf\purple{↬}$Time is 10 sec

Then,

$\sf\purple{↬}$Time travels for 30 sec

$\sf\purple{↬}$Constant speed

Then,

$\sf\purple{↬}$Final speed is 0m/s

$\sf\purple{↬}$retardation is 4m/s²

$\boxed{\rm{\red{To\:Find\longrightarrow }}}$

$\sf\purple{↬}$Total Distance covered

$\boxed{\rm{\orange{Formula\:used \longrightarrow }}}$

$\sf{s= ut+\dfrac{1}{2}at^2}$

$\orange\bigstar\:\rm{\gray{\overbrace{\underbrace{\pink{v^2 - u^2 = 2as\:}}}}}$

where,

• s=distance
• u=initial speed
• a=acceleration
• t=time
• v=final peed

$\boxed{\rm{\green{Solution \longrightarrow }}}$

☛Initially,

by using 2nd equation,

$\sf{s= ut+\dfrac{1}{2}at^2}$

$\sf{s= 0\times 10+\dfrac{1}{2}\times 2 \times 10^2}$

$\sf{s= \dfrac{1}{2}\times 2\times 100}$

$\sf{s=100m}$

here distance s = 100m

 

☛Then,

v=20

t=30

Traveling for 30sec with constant speed,

so here distance covered is,

$\sf{s' =vt}$

$\sf{ s' = 20\times 30= 600m}$

Distance s'= 600m

Distance s'= 600m 

☛Now,

Travelling due to retardation,

By using 3rd equation,

$\orange\bigstar\:\rm{\gray{\overbrace{\underbrace{\pink{v^2 - u^2 = 2as\:}}}}}$

$\sf{(0)^2 - (20)^2 = 2 \times (- 4) \times s''}$

s'' = 50 m

 

Now, the total distance covered,

S= s+s'+s''

S= (100 + 600 + 50) m

S=  750 m

Hence, Distance covered by the particle is 750 m.