Question

The velocity of an object with a mass of `5 kg` is given by `v(t)= 2 t^2 + 9 t`. What is the impulse applied to the object at `t= 7`?

Answer 1

`805Ns`

Explanation:

Step 1:
We know,
`v(t)=2t^2+9t`

Putting `t = 7`,
`v(7)=2(7)^2+9(7)`
`v(7)=98+63`
`v(7)=161m/s` ---------------- (1)

Step 2:
Now, `a=(v_f-v_i)/(t)`
Assuming the object started from rest,

`a=(161m/s-0)/(7s)`
`a=23m/s^2` ------------------- (2)

Step 3:
`"Impulse" = "Force"*"Time"`
`J=F*t`
`=>J=ma*t` ---------- (`because` Newton's 2nd law)

From (1) & (2),
`J=5kg*23m/s^2*7s`
`=805Ns`

Answer 2

`805 Kgms^-2`

Explanation:

Impulse is defined as change in momentum,i.e `m(v-u)`
Where, `v` is the final velocity and `u` is the initial velocity of mass `m`

Now,given velocity-time relationship is `v=2t^2+9t`

So,`mv=5(2t^2+9t)=10t^2+45t`

So,`m(v_7 -v_0)=10(7)^2+45×7-(0+0)=805Kg ms^-2`