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 $t= 7$ ?

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Answer 1 · 2018-03-13T16:18:59

$805Ns$

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 · 2018-03-13T16:28:29

$805 Kgms^-2$

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

The velocity of an object with a mass of 5 kg5 kg is given by v(t)= 2 t^2 + 9 t v(t)= 2 t^2 + 9 t . What is the impulse applied to the object at t= 7 t= 7 ?