Question

Can some answer 7 a and b please

Answer 1

a.After
`15` second Mr Comer's speed
`=`
`1.7 m/s`

b.Distance travelled by Mr.Comer in
`15` seconds
`=`
`18.75\ m`

Step-by-step explanation:

a. Lets recall our first equation of motion
`V_(f) =V_(i) + at`

Now we know that
`V_(i) = 0.8m/s` ,
`a = 0.06\ m/s^2` and

`t = 15 \ sec`

Plugging the values we have.

`V_(f) =V_(i) + at`

`V_(f) =0.8 + 0.06 * 15`

`V_(f) =0.8+ 0.9`

`V_(f) =1.7 m/s`

Then Mr.Comer's speed after
`15` sec
`=`
`1.7ms^-1`

b.

Lets find the distance and recall our third equation of motion.

`V_(f) ^2-V{i}^2 = 2as`

So
`s =` distance covered.

Dividing both sides with 2a we have.

`\frac{V_(f) ^2-V{i}^2}{2a} = 2as`

Plugging the values.

`((1.7)^2-(0.8)^2)/(2*0.06) = 2as`

`s= 18.75\ m`

So Mr.Comer will travel a distance of
`s= 18.75\ m`.