Question

A particle is moving along a staright line decelerates uniformly from 40cm/s to 20cm/s and then has a constant acceleration of 20m/s2 during the next 4sec. The average speed during the whole time interval in cm/s is
a-57
b-140
c-86
d-43

Answer 1

@kropot72

Answer 2

In your question:
a) Should "......acceleration of 20m/s2........" have been written as "...acceleration of 20cm/s^2......" ?
b) The time taken to decelerate uniformly from 40cm/s to 20cm/s has not been given. Please check your posted question against the original question.

Answer 3

Time is 5 sec
and am sorry it is cm/s^2

Answer 4

The average speed during the deceleration is given by:
\[\frac{40+20}{2}=30\ m/s\]
The distance traveled during the deceleration is
\[30\times5=150\ cm\]
The final speed at the end of the acceleration is
\[20+(4\times20)=100\ m/s\]
The average speed during the acceleration is
\[\frac{20+100}{2}=60\ m/s\]
The distance traveled during the acceleeration is
\[60\times4=240\ cm\]
The total distance traveled is 150 + 240 = 390 cm in a total of 5 + 4 = 9 seconds.
Now use the formula
\[speed=\frac{distance}{time}\]
to find the solution.

Answer 5

For average speed which formula do you use ? Can ya right it in general form and Thank you So MUCH ^_^

Answer 6

YAY \O/ I got the answer its D . ^_^ Thanks a bunch :D

Answer 7

Your choice is correct! Good work :)

Answer 8

but wait please tell me the general formula for average speed So it can help me

Answer 9

in future *

Answer 10

40+20/2=30 m/s
total Acceleration/time = average speed ???/

Answer 11

For motion in a straight line, when the speed changes uniformly with time, the average speed in any time interval equals one-half the sum of the speeds at the beginning and at the end of the interval. Therefore the average speed v-bar between t = 0 and t = t is
\[\bar {v}=\frac{v _{0}+v}{2}\]

Answer 12

oh got it . Thank you so much ya are superb . Such a awesome explanation ^_^

Answer 13

So , using this formula for the deceleration time interval:
\[v _{0}=40\ m/s\]
\[v=20\ m/s\]
\[\bar {v}=\frac{40+20}{2}=30\ m/s\]

Answer 14

And isn't it cm/s instead of m/s in average speed ??

Answer 15

My bad :( Good spotting!

Answer 16

Thank you so much ^_^

Answer 17

You're welcome :)