Question

A car travels on a straight road with a velocity of 9 m/s. Driver applies break and produces a retardation of 3 m/s². Find the distance covered by the car in 2 sec. and after 5 sec. after applying the break.

Answer 1

Given,

u = 9 m/s

v = 0 m/s

a = -3 m/s^2

t1 = 2 s

v2 -u2 = 2as

s = ut+1/2a t1^2

  = (0)(2) + 1/2(3)(2)^2

  = 1/2(3)(2)(2)

  = (3)(2)

  => 6 m

t2 = 5s

s = ut+1/2a t2^2

  = (0)(5) + 1/2(3)(5)^2

  = 1/2(3)(5)(5)

  = 1/2 (75)

  => 37.5 m

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Answer 2

Step-by-step explanation:

hlo!! Given ,

u=9m/s

v=0m/s

a=-3m/s^2

t1 =2s

V2-U2=2as

S=at+1/2a t1^2

=(0) (2)+1/2 (3)(2)^2

=1/2 (3) (2) (2)

=(3) (2)

=6cm

t2=5s

S=it+ 1/2a t2^2

=(0) (5) +1/2 (3) (5)^2

=1/2 (3) (5) (5)

=1/2 (75)

=37.5m

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