Question

The speed of a vehicle is reduced with a constant acceleration from 72km/h to 18

Answer 1

Answer:

The correct answer will be "1477.84 N".

Explanation:

Given that,

Mass,

m = 1.6 mg

or,

   = 1600 kg

Initial velocity,

u = 72 km/h

  = $72\times \frac{5}{18} \ m/s$

  = $20 \ m/s$

Final velocity,

v = 18 km/h

  = $18\times \frac{5}{18}$

  = $5 \ m/s$

Covered distance,

s = 250 m

By using the below relation, we get

⇒  $v^2=u^2+2as$

On putting the values, we get

⇒  $(5)^2=(20)^2+2\times a\times 250$

⇒      $a=-0.75 \ m/s^2$ (shows the deceleration)

Slope will be given as 1 in 25, then

⇒  $Sin \theta=\frac{1}{25}$

           $\theta=2.3^{\circ}$

hence,

As we know,

⇒  $\Sigma F=ma$

or,

⇒  $Braking \ force+350-mgSin\theta=ma$

⇒  $Braking \ force=ma+mgSin\theta-350$

On substituting all the values, we get

⇒                           $=1600(0.75+1600\times 9.81 Sin(2.3^{\circ})-350$

⇒                           $=1477.84 \ N$