Question

plz ans my question.​

Answer 1

Answer:

1.OBJECT B HAS HIGHEST ACCELERATION.

INITIAL VELOCITY =25 M/S

FINAL VELOCITY =70 M/S

TIME TAKEN = 40 -10=30 SECONDS

ACCELERATION:

$\frac{v - u}{t} = \frac{70 - 25}{30} = \frac{45}{30} = 1.5 \: m \: {s}^{ - 2}$

2.●ACCELERATION OF OBJECT A:

$\frac{25 - 10}{30} = \frac{15}{30} = 0.5m \: {s}^{ - 2}$

●ACCELERATION OF OBJECT B IS 1.5 M/S^2 .

●ACCELERATION OF OBJECT C:

$\frac{15 - 30}{30} = \frac{ - 15}{30} = - 0.5 \: m {s}^{ - 2}$

3.WE CAN IDENTIFY RETARDATION BY OBJECT C BECAUSE WE SEE THE FINAL VELOCITY ACHIEVED IS SMALLER THAN INITIAL VELOCITY.

FINAL VELOCITY =15 M/S

INITIAL VELOCITY =30 M /S

YOU KNOW 15 IS SMALLER THAN 30.

4.TOTAL TIME TAKEN BY ALL THREE OBJECTS =

10+20+30+40=100 SECONDS

●NOW TOTAL VELOCITY OF OBJECT A =

10 +15+20+25=70 M/S

DISTANCE = V×T=100 ×70=7000 M

● TOTAL VELOCITY OF OBJECT B=

25+40+55+70=190 M /S

DISTANCE = V×T=190×100=19000 M

● TOTAL VELOCITY OF OBJECT C:

30+25+10+15=90 M/S

DISTANCE = V×T=100×90=9000 M

NOW SUM OF DISTANCE BY ALL THREE OBJECTS:

7000 +19000+9000=35000 M.