# 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 ?

Mar 13, 2018

$805 N s$

#### Explanation:

Step 1:
We know,
$v \left(t\right) = 2 {t}^{2} + 9 t$

Putting $t = 7$,
$v \left(7\right) = 2 {\left(7\right)}^{2} + 9 \left(7\right)$
$v \left(7\right) = 98 + 63$
$v \left(7\right) = 161 \frac{m}{s}$ ---------------- (1)

Step 2:
Now, $a = \frac{{v}_{f} - {v}_{i}}{t}$
Assuming the object started from rest,

$a = \frac{161 \frac{m}{s} - 0}{7 s}$
$a = 23 \frac{m}{s} ^ 2$ ------------------- (2)

Step 3:
$\text{Impulse" = "Force"*"Time}$
$J = F \cdot t$
$\implies J = m a \cdot t$ ---------- ($\because$ Newton's 2nd law)

From (1) & (2),
$J = 5 k g \cdot 23 \frac{m}{s} ^ 2 \cdot 7 s$
$= 805 N s$

Mar 13, 2018

$805 K g m {s}^{-} 2$

#### Explanation:

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

Now,given velocity-time relationship is $v = 2 {t}^{2} + 9 t$

So,$m v = 5 \left(2 {t}^{2} + 9 t\right) = 10 {t}^{2} + 45 t$

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