If a car moving at 50km/h skids15m with locked brakes, how far will car skid with locked brakes at 150km/h?

Answer 1

For 50 km/h skids 15m with brake

v^2 = u^2 - 2as
0 = (50x10^3 / 3600)^2 - 2a(15)
(50x10^3 / 3600)^2 = 2a(15) ---(1)

For 150 km/h,

v^2 = u^2 - 2as
0 = (150x10^3 / 3600)^2 - 2as
(150x10^3 / 3600)^2 = 2as ---(2)

(2)/(1)

(150 / 50)^2 = s/15
s = 135 m

Answer 2

The initial velocity is 50 km/h and the final velocity is 0. So, let us use the equation, v^2 - u^2 = 2 a.s where s is the acceleration and s is the distance travelled.

Here we have 0 - 50 X 50 = 2. -a . 15 ( we use - sign before a to denote deceleration). The negative sign gets cancelled out.

50 X 50 = 30 a or a = 2500 / 30 = 250/3

Now let us calculate the second equation:

0 - 150 X 150 = 2. 250 .-s / 3

Again ignoring the negative sign on both sides,

150 X 150 = 500s/3

s = 3.150.150/500 = 3.3.50.150/500 (we just expressed 150 as 3.50 for simplification)

= 9.150/10 = 9.15 = 135 m

Please note that we have not used a calculator anywhere in the above solution.

Answer 3

here, in this concern, the conception must be that the frictional tension continues to be an analogous and so the deceleration produced additionally to be an analogous. by equations of action, v^2 = u^2 + 2as (notice that ingredient is absent in this equation and that too time isn't given interior the priority) because of the fact the case is with deceleration, ie destructive acceleration and intensely final speed v =0 because of the fact the vehicle is composed of relax, the expression will become 2as = u^2 consequently the relation between s and u is: s is immediately proportional to the sq. of the fee. subsequently, s2/s1 = (u2/u1)^2 advice: Which one to be discovered would desire to be interior the left side numerator. here s2 the skidding distance is needed. particularly straightforward manipulation. So s2 = s1*(u2/u1)^2 it particularly is how we would desire to grant the attitude particularly collectively as writing any front examination attending purpose style questions. Now plugging the given values and simplifying we get, s2 =15 * (a hundred and fifty/50)^2 = 15 * 9 = one hundred thirty five m

Answer 4

If it is a straight linear equation then-:

150 / 50 = 3

3 x 15 = 45 metres

Answer 5

Since we expect that their deceleration rates to be equal, we can use ratio and proportion.

Since a = v^2/(2d), then

v1^2/(2d1) = v2^2/(2d2)
50^2/(2*15) = 150^2/(2*x)
x = (150^2/50^2)*15
x = 135 m.