A 2200 kg car doubles its speed from 50 km/hr to 100 km/hr. By how many times does the kinetic energy from the car's forward motion increase?

Answer:4

Explanation:m=2200, u=50, v=100

The unit for speed is in m/s we have to conver to m/s

For ke1=50km/h to m/s 50*1000/3600=13.89

So keep=1/2mv^2

1/2 (2200) 13.89^2 = 212225.31

Ke2=1/2mv^2=100*1000/3600=27.78

1/2mv^2 = 1/2 (2200) 27.78^2

=848901.24 ... soke2/ke1 = 848901.24/212225.31=4

That d increase in ke

Answer: 4

Explanation:

Given that:

Mass of car M = 2200 kg

Initial speed Vi = 50 km/hr

Final speed Vf = 100 km/hr

Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the speed (v) by which it moves

i. e K. E = 1/2MV^2

So, when traveling at 50 km/h

KE = 1/2x 2200kg x (50km/h) ^2

KE = 0.5 x 2200 x 2500

KE1 = 2750000J

So, when traveling at 100 km/h

KE = 1/2x 2200 x (100 km/h) ^2

KE = 0.5 x 2200 x 10000

KE2 = 11000000J

Thus, the number of times kinetic energy increases is obtained by dividing KE2 by KE1

i. e 11000000J / 2750000J

= 4

Thus, the kinetic energy from the car's forward motion increase 4 times