Question

A car moving with speed of 50 m/s ,can be stopped by brakes after at least 6m. if the same car is moving at a speed of 100m/s, the minimum stopping distance is

Answer 1

ANSWER

If final velocity is zero and acceleration is −ve then.

s= 2a/u ^2

s∝u ^2

So, if the velocity is doubled than stopping distance will become 2

2

=4 =6×4

=24 m.

Answer 2

Answer:

let's suppose

that car moving with the speed of 50 m/s is CAR A...and the car with the speed of 100m/s is CAR B.

SO, we know that

CAR A:

speed: 50m/s

stops at: 6m

CAR B:

speed: 100m/s

stops at: xm

(STEP 1)

Speed. Stops at.

50 m/s. 6 m

(apply unitary

method--

cross multiply

them)

100m/s. x m

(STEP 2)

50*X = 100*6

(STEP 3)

50x = 600

(STEP 4)

x= 600÷ 50

x= 12 ANSWER

Hope this helps!