Question

plz view this?

Answer 1

plz answer with steps Q 135 1
36

Answer 2

To solve this you will need to know 2 kinematics equation:

V= u +at , where v= final velocity , u = initial velocity, a = acceleration, t= time

s= ut + (1/2) a t^2 where, s = distance travelled.

Given that the ratio for the distance travelled in the 5th second and the distance travelled in the 7th second is 31/27

Form an equation,

(distance travelled in 5s) - (distance travelled in 4s) 31
-------------------------------------------- = ----
(distance travelled in 7s) - (distance travelled in 6s) 27


Using the kinematics equation stated above, fully from the equation then

\[\frac{(5u + \frac{25}{2}a) - (4u+\frac{16}{2}a)}{(7u+\frac{49}{2}a)-(6u+\frac{36}{2}a)}= \frac{31}{27}\]

Fully expand,
\[\frac{u+\frac{9}{2}a}{u+\frac{13}{2}a}= \frac{31}{27}\]

Cros multiply,
\[27u+\frac{243}{2}a= 31u+\frac{403}{2}a\]

Simplyfy,
\[u=-40a\]

With the above equation, you can solve the question already.

When the body will stop in time, is when v = 0

using equation, v=u+at

0= u+at (substitute the above u=-40a
0=-40a +at (cancel out the a's)
0=-40 +t
t= 40s (Ans)

Answer 3

hey the answer of( q 135) is 20s plz can u recheck ur ans

Answer 4

haha sry didnt know there was one more question.

Anyway, since the values are the same, you will arrive at the above answer t= 40s, meaning that the body will stop at 40s. now you want to find the distance travelled in the 40s. No matter how hard you try, you will not get the answer, because u have no other values to use. You need either the value for u(initial velocity) or a (the acceleration) to solve for the distance travelled. so my answer would be (d) data insufficient

Answer 5

yes u r correct it is data iinsufficient but can u check........ 135 t=20s

Answer 6

lol if u went through my workings, u will know i made a calculation error.

u=-20a.

Answer 7

ok