Q)A CAR IS TRAVELLING ON A STRAIGHT ROAD.THE MAX VELOCITY THE CAR CAN ATTAIJN IS 24 M/S.THE MAX ACCELERATION AND RETARDATION IT CAN ATTAIN IS 1 AND 4 RESPECTIVELY.THE SHORTEST TIME THE CAR TAKES FROM REST TO REST IN A DISTANCE OF 200M IS....ANS)22.4S

|----------------|---------------|

|<------x----->|<---200-x--->|

let us assume distance travelled during accn is x and time taken is t1 nd distance travelled during retardation is 200-x and time taken is t2......then for accn we can write newtons eqution as---

v=u+at1

u=0 initially at rest

thus v=t1......a=1m/s²------(1)

now applying newtons second law of motion .....

s=ut+1/2(at²)

x=(1/2)t1²....as u=0,a=1 .....

v=t1=√2x........(2)

now considering the retardation path and newtons (3) law of motion

v²-u²=2as              v=u+at.......v=0.....u=√2x....a=-4........t2=(√2x)/4.....(3)

as we know initial velocity will be the final velocity of acceleration path and final velocity will be zero thus

-u²=-8(200-x)....as a=-4m/s²

(√2x)²=2(800-4x)....as u=√2x

x=800-4x

5x=800.................x=160..........t1=√2*160(from eqn 2)

thus t1=8√5

t2=2√5....from eqn 3

total time=t1+t2=10√5=22.5sec......answer

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similarly, you can chech other case that v1 reaches 24m/s dur. acc. but time reqd. will come to be more.