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#