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 ?

2 Answers
Mar 13, 2018

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

Mar 13, 2018

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