Question

Two cars leave with a time gap of 1 minute from
the same point. They move with an acceleration of
0.2 m/s^2. How long after the departure of the
second car, does the distance between them
become equal to three times its value, when the
second car just starts?​

Answer 1

Answer:

Answer:

Since car A Starts 1 minute before the car B.

Explanation:so we have to calculate first the distance and speed of car A.

For car A — u=0,a=0.2m/s^2,t=1minute or60 second.

To find distance formula is “S=UT+1/2AT^2”.

s=0×t+1/2×0.2×60×60

S=0.2×60×30

S= 360meter.

So distance travelled by car a in 60 seconds is 360 meter.

To calculate speed of car A formula is

SPEED=TOTAL DISTANCE /TOTAL TIME.

DISTANCE =360 M, TIME =60 SECONS.

SPEED OF CAR A=360/60

SPEED =6METER/SECOND..

Since time of car a and car b both are same so distance travelled and speed of both car are same.

SO FOR DIFFERENCE OF DISTANCE BETWEEN CAR A AND CAR B IS THRICE AND WE HAVE TO FIND THE TIME AT WHICH THE DIFFERENCE IS THRICE...

FOR THAT WE MULTIPLY THE DISTANCE AND SPEED OF CAR A WITH 3 BECAUSE CAR A IS IN FORWARD OF CAR B.

DISTANCE OF CAR A = 360×3=1080.

SPEED=6×3=18M/S.

AS WE KNOW THAT TIME =DISTANCE /SPEED.

SO TIME AFTER WHICH THE DISTANCE BETWEEN CAR A AND B IS BECOME THRICE = 1080/18=60SECOND OR 1 MINUTE.

SO AFTER I MINUTE OF DEPARTURE OF CAR B THE DISTANCE DIFFERENCE BETWEEN CAR A AND CAR B IS BECOME THRICE.......

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