114 - Kinematics Questions Answers
a particle , with an inital velocity v0 in a plane , is subjected to a constant acceleration in the same plane . then in general ,
the path of the particle would be
1) circle
2) ellipse
3) parabola
4) hyperbola
parabola
for understanding this make components of v parallel and perpendicular to a
a train is moving with a velocity of 30m/s. When brakes are applied , it is found that the velocity reduces to 10m/s in 240m.
When the velocity of the train becomes zero , the total distance travelled is??
270m
by v2 = u2-2as
100 = 900-2*a*240
or 800 = 480a
or a = 80/48 = 10/6 = 5/3
so total distance upto rest is u2/2a = 900*3/2*5 = 270m
angle b/w vectors (i^+j^)and(j^+k^) is
1) 60o
2) 90o
3) 300
4) 450
use the formula A.B = |A||B|cosθ
A at road of 200km & B at 100km road. A with vel 20m/s & B with 7.5m/s . Then wat will b d time when B overtake A.
1) 24s
2) 40s
3) 80s
4)nothing can be predicted
data are irrelevent
QUES A MAN RUNS AT A SPEED OF 4M/s TO OVERTAKE A STANDING BUS . WHEN HE IS 6m BEHIND THE DOOR (AT t=0) , THE BUS MOVES FORWARD & CONTINUES WITH A CONSTANT ACCELERATn OF 1.2m/s2 . the MAN SHALL GAIN THE DOOR AT TIME t EQUAL TO
A) 5.2s
b) 4.3s
c)2.3s
d) the man shell never gain the door.
ans 4.3s , Sir ,on solving I 'm getting both 4.3 & 2.3 s ( when we solve sq root) so here i want to know why not the ans is 2.3s. AMU 2012
After 2.3 sec. the man overtake the door first, in this time velocity of bus will be less than that of man so it is not possible to catch the bus then after 4.3 sec. the bus door will again reach to the man, in this time velocity of bus and man are same, so relative velocity is zero and catching of bus is possible.
A BODY STARTS WITH INITIAL VELOCITY 30m/s & A RETARDATION OF 4m/s2 . FIND THE DISTANCE TRAVELLED BY THE BODY IN 8th SECOND.
ANS 1m
by v = u - at
0 = 30 - 4t implies t = 7.5
it means velocity will become 0 in 7.5 sec.
distance covered in 7.5 sec. = 30*7.5 - 1/2*4*(7.5)2 = 225 - 225/2 = 225/2
distance covered in 7 sec. = 30*7 - 1/2*4*(7)2= 210 - 98 = 112
so distance covered in first half of eighth sec. = 112.5 - 112 = 0.5 m
distance covered in last half of eighth sec. = 0*0.5 + 1/2*4(0.5)2= 0.5
so total distance covered in eighth sec. = 0.5 + 0.5 = 1 meter
A TOY CAR TRAVELS FROM A TO B AT A CONST. SPEED OF 20 km/h & IMMEDIATELY RETURNS TO A AT A CONST. SPEED v. IF THE AVG. SPEED OF THE CAR IS 24 km/h . THEN v IS = ?
total distance = d+d = 2d km.
total time taken = d/20 + d/v hours
so avg. speed = 2d/[d/20 + d/v]
or 24 = 2/[1/20 + 1/v]
or 12 = 1/[1/20 + 1/v]
or 1/20 + 1/v = 1/12
or 1/v = 1/12 - 1/20
or 1/v = 2/60
or v = 30
A hunter aims his gun at a monkey sitting on a tree at the instant bullet leaves the gun the monkey drops down . WILL THE BULLET HIT THE MONKEY ?
yes because monkey and bullet both are falling with the same gravity
IN QUESTn < B/W 2 NON ZERO VECTOR A &B 120* RESULTANT GIVEN IS GREATER WHILE ACC. TO IT SHD BE LESS SO I REQUEST YOU SIR TO AGAIN CHECK . QUES FROM HC VERMA
i am confirmed, option B will be correct answer
Q A CURVE IS REPRESENTED BY y=sinx IFx IS CHANGED FROM PIE/3 TO PIE/3+PIE/100, FIND APPROX. CHANGE IN x?
Q A CAR IS GOING DUE NORTH AT A SPEED =50km/h IT MAKES A 90* LEFT TURN WITHOUT CHANGING SPEED THE CHANGE IN VEL. OF CAR IS=? (ANS 70km/h towards south west ) i want to know that here we are just changing the directn so only directn. shd. b change but here magnitude is also changing HOW?
FLUIDS
Q A BAROMETER KEPT IN N ELEVATOR READS 76cm WHEN IT IS AT REST . IF ELEVATOR GOES UP WITH INCREASING SPEED READING WILL BE <76 cm . IS REASON IS THAT PRESSURE IS DIRECTLY PROPTn TO HEIGHT ?
I think we have to caculate change in y
since y = sin x
so dy = cosx dx = cosπ/3* π/100 = π/200
the initial velocity of car = 50j and after turn velocity = -50i here i and j are the unit vectors along east and north direction
so change in velocity = -50i - 50j
its mod = 50 √2 = 70
Because in barometer the extra Hg level will experience a downward force so Hg level will decrease.