IIT JEE , AIEEE Forum - Physics , Mechanics

piyushsahani

1 .Can a body has zero velocity and still be accelerating? If yes then give example and if no then explain why?

2. Car covers the first half of the distance between two places at the speed 40km/hr and the second half at the speed of 60 km/hr .what is the average speed of the car?

3. A particle moves along x- axis in such away that its co-ordinates x varies with the time?t? according to the expression:

x = (2- 5t + 6t ²⁾ m

What is the initial velocity of the particle?

4. The velocity of the particle at the instant?t? is 10 m/s. After 5 s the velocity of the particle is 20 m/s .then what is the velocity of the particle 3s before

5. A boat takes two hours to travel 8 km and back in Still Water Lake. With water velocity 4 km/hr. The time taken for going upstream 8km and coming back is _________________ minutes.

6. If the velocity of the particle is given by u = root of (180 -16x) m/s its acceleration will be __________________ .

joyfrancis

q1)
answer is yes.
Example
When we throw a body directly upwards the velocity at the highest point is zero and the acceleration for the entire motion is g.The velocitycan be zero only momentarily.

nunoxic

for the last one ,

6. If the velocity of the particle is given by u = root of (180 -16x) m/s its acceleration will be __________________

simply differentiate it with respect to t

thus ,
du/dt = [1/ 2 root 180 - 16x ] into -16

a = [1/ 2 root 180 - 16x ] into -16

becuz
when x = displacement
dx/dt - velocity
dv/dt - acceleration (which is also written as d2x/dt2)

Rate If Useful

nunoxic

5. A boat takes two hours to travel 8 km and back in Still Water Lake. With water velocity 4 km/hr. The time taken for going upstream 8km and coming back is _________________ minutes.

i dont quite understand this question
in still lake, the speed of boat is 8/2 = 4 kmph
so when going upstream, resultant speed will be
= speed in still water - speed of river
but thats 0
y will the boat ever move ?
i seem to be making a mistake i think.......

bhuvana89

ans 2)
let total distance = L
For first half L/2 , time taken = L/2*40 = T1
FOR second half , time taken = L / 2 * 60 = T2
AVG SPEED = TOTAL DISTANCE/ TOTAL TIME TAKEN
= L / T1 + T2
= 45 KM /HR

bhuvana89

ANS 3)

x = 2 - 5t + 6t ^2

diffentiating

v = -5 + 12 t

put t = 0

initail velocity = -5 m/s

bhuvana89

ans 4 )

for first case
v= 20 m/s
u = 10 m/s
t =5 sec
use v= u + at to find acceleration
solving a = 2 m/s^2

now
v= 10 m/s
a= 2 m/s^2
t = 3 sec
u = ??

use v = u + at again to find u

ans is 4m/s

nivedh_89

Bhuvana i am afraid that ur 2nd answer is wrong......ans is 48km/hr...........
t1 = x/40...
t2 = x/60....
total time = t1+t2.... = x/24....therefore....x/t = 24
But...Total distance = 2x.....

therefore....total distance/total time = 2x/t = 2*24 = 48 = average speed.......

bhuvana89

ans 5 )
let the velocity of man in boat =u
velocity of river = v

in still water'
total time of journey = 2 hrs
=> 2 = 8/u + 8/ u ( using time = distance/speed )

solving we get u= =8 km/hr

with water velocity
time taken 4 upstream = distance / u-v
time taken for downstream = distance / u+v

so t = 8/12 + 8/4
= 8/3 hr

= 160 minutes...

nivedh_89

REST ANSWERS ARE GIVEN CORRECTLY BY OTHERS.............

bhuvana89

ya it is 48 km/hr for ans 2)

sorry typing mistake.....!!!!!!!!!!!!!!!!!!!!

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