Question

A point moves along a straight line in such a way
That after t seconds its distance from origin is s=2t^2+3t meters
i)find average veleity of the points between t=3, t=6seconds
ii) Find instantaneous velocities at t=3, t=6 seconds
​

Answer 1

Answer:

Answer:

∫

6

−

3

v

(

t

)

d

t

=

103.5

Step-by-step explanation:

Explanation:

The area under a velocity curve is equivalent to the distance covered.

∫

6

−

3

v

(

t

)

d

t

=

∫

6

−

3

−

t

2

+

3

t

−

2

X

d

t

=

−

1

3

t

3

+

3

2

t

2

−

2

t

∣

∣

∣

6

(

−

3

)

=

(

−

1

3

(

6

3

)

+

3

2

(

6

2

)

−

2

(

6

)

)

−

(

−

1

3

(

−

3

)

3

+

3

2

(

−

3

)

2

−

2

(

−

3

)

)

=

114

−

10.5