Question

a motorcar of mass 1200kg is moving along a straight line with uniform velocity of 90km/h.its velocity is slowed down to 18km/h in 4 sec.its velocity is slowed down to 18km/h in 4 sec by an unbalanced external force. Calculate in magnitudes ion of the force required?

Answer 1

given,
u=90km/h=90×1000/3600m/s=25m/s
v=18km/h=18×1000/3600m/s=5m/s
t=4sec
so, a=(v-u)/t=(5-25)/4=-5m/s²
Force, F=m×a=1200×(-5)=-6000N
Magnitude of external force is 6000N opposite to the direction of velocity.

Answer 2

Given :

mass (m) = 1200 kg

Initial velocity (u) = 90 km/h = 25 m/s

Final velocity (v) = 18 km/h = 5 m/s

Time (t) = 4 sec

[ To convert km/h into m/s we have to multiply 5/18 ]

(i) acceleration (a) = $\bf \frac{v-u}{t}$

                     

                              = $\bf \frac{5-25}{4}$

                           

                              = $\bf \frac{-20}{4}$

                              = -5 m/s

Now,

We got (a) = -5 m/s which indicates that the velocity decreases

(ii) Change in momentum

  =  Final momentum - Initial momentum

  = $\bf mv - mu$

  = $\bf m(v-u)$

  = $\bf 1200(5-25)$

  = $\bf 1200(-20)$

  = -24000 kg - m/s

(iii) Magnitude of the force required

  = Rate of change of momentum

  = $\bf \frac{Change\:\: in \:\:Momentum}{Time}$

  = $\bf \frac{-24000}{4}$

  = - 6000 N