Question

Initially a car is moving with speed of 20 ms and
covers a distance of 100 m before it comes to rest.
On making the initial speed of car 40 ms-
keeping the acceleration same, the distance
travelled by car before it comes to rest is
(1) 200 m
(2) 300 m
(3) 400 m
(4) 800 m

The answer is the 3rd option plz give explanation..

Answer 1

Explanation:

in first case,

u=20m/s and v=0(it comes to rest)

s=100m so by using v^2 -u^=2as we get,

-400 =200a, so acceleration is -2.

in second case,

u=40m/s and v=0,

a= -2(as acceleration remains constant)

so by using v^2-u^2=2as,

-1600=2×-2×s

-1600= -4×S

S=400m

Answer 2

Answer:

400m

Explanation:

First case,

Given,

u=20m/s

s=100m

v=0

Now third equation of motion,

v^2=u^2+2(a)(s)

0=(20)^2+2(a)(100)

0=400+2a(100)

-2a(100)=400

-2a=400/100

-2a=4

a=4/-2

a=-2m/s^2

Now in the second case,

Given,

u=40m/s

a=-2m/s^2

v=0

By 1st equation of motion,

v=u+at

0=40+2(t)

-2t=40

t=40/-2

t=20s

Now by using 2nd equation of motion,

s=ut+1/2(a)(t)^2

s=40(20)+1/2(-2)(20)^2

s=800+(-1)(400)

s=800-400m

s=400m.

Hope it helps you!

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