Question

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Answer 1

Answer:

Explanation:

i) during first two seconds

v/t =a so slope graph is acceleration

slope of graph =tan theta

y/x= 4.6 /2= 2.3 m/s square

iii) similarly last two seconds

but it deaccelerates so it will be in negative sign

tan theta= -4.6/2=-2.3 m/s square

ii) as slope of graph constant , tan theta =dy /dx

from differentiation

acceleration =0

Answer 2

Hello!

Nice Question

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For the first two seconds --

Acceleration = Change in velocity/Time

Acceleration= $\frac{(v-u)}{t}$ (IMPORTANT)

By observing the graph--

v=4.6 m/s

u=0 m/s

t= 2 secs

Hence, a= $\frac{4.6}{2}$ = 2.3 m/s^2

For the last two seconds--

v= 0 m/s

u= 4.6 m/s

t= 2 secs

a= $\frac{(v-u)}{t}$

a= $\frac{0-4.6}{2}$

a= -2.3 m/s^2 (Negative sign indicates deceleration)

Between 2nd and 10th seconds--

Since the line is a straight line parallel to the time axis

The velocity is constant (velocity does not increases or decreases)

Hence, Acceleration=0 m/s^2

Hope that helps you

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