A car of mass 500kg reduces speed from 100km/hr to 30km/hr in 20 seconds.?

Question

A car of mass 500kg reduces speed from 100km/hr to 30km/hr in 20 seconds. Calculate the breaking power required to make this change in speed.

SHOW ALL WORKING OUT

1ST F=MASS X ACCELERATION

2ND WORK DONE = FORCE X DISTANCE

3RD POWER =WORKDONE/TIME

Answer 1

u=100km/hr =27.77 m/s

v= 30km/hr =8.33 m/s

using,

v=u +at

we get,

a= -1 m/s^2 (approx.)

1ST F=MASS X ACCELERATION

F=500 * 1 N

=500 N



to find the distance,use,

s= ut + 1/2 at^2

solving this,we get,

s=355.4 m (approx.)

2ND WORK DONE = FORCE X DISTANCE

W=500 * 355.4 J

=17777 J(approx.)

3RD POWER =WORKDONE/TIME

P=17777/20

=888.85 watt

Please verify the answer as I have calculated just once

Answer 2

This should get you going.  Remember to keep track of the signs.

[1st]

∆v = v(f) - v(i) , [final velocity - initial velocity]

?km/hr * 1000m/1km * 1hr/3600s = ?m/s

a = ∆v/∆t

F = m * a

[2nd]

d = v(i) * t + (1/2) * a * t^2

W = F * d

[3rd]

P = W / t

Answer 3

<< 1ST F=MASS X ACCELERATION >>

Acceleration = (100 - 30)*(1000/3600)/20

Acceleration = 0.972 m/sec^2

Therefore,

Force = 500 * 0.972 = 486 N

<< 2ND WORK DONE = FORCE X DISTANCE >>

Work = 486 * S

To determine the distance travelled by the car in 20 seconds,

use the formula

S = VoT - (1/2)aT^2

where

S = distance travelled in 20 sec.

Vo = initial velocity = 100 kph = 27.78 m/sec.

a = 0.972 (as calculated above)

T =20 sec.

Substituting appropriate values,

S = 27.78(20) - (1/2)(0.972)(20^2)

S = 555.60 - 194.40

S = 361.20 meters

Therefore,

Work done = (486)(361.20) = 175,543.20 joules

<< 3RD POWER =WORKDONE/TIME >>

Power = 175,543.20/20

Power = 8777.16 joules/sec = 8777.16 watts = 8.77716 KW