Question

with the help of graph solve the question step by step​

Answer 1

Explanation:

a) distance between t=0s and t=5s

since area under a v-t graph gives us distance as v×t=s

so, s(0-5)s=Ar(trap.AOBD)

=1/2(5+3)×20

=80m

b) distance between t=0 to t=10 s

= Ar(trap.AOBC)

=1/2(10+3)×20

=130m

c) acceleration for OA,

u=0m/s

v=20m/s

t=2 s

so a=(v-u)/t

=20/2

=10m/s^2

acceleration for BC,

here, u=20m/s

v=0m/s

t=5s

so a=(v-u)/t

=0-20/5

=-20/5

=-4m/s^2

I hope it helps you

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