Question

Help me with the attached file...:D

Answer 1

The car covers D distance in 20 seconds
from t=20 to t=60 the car's speed is u m/s
so D meters it'll cover in
\[Time\ t=\frac{D}{u} seconds\]
So total time taken to cover 2D meters= \[20+\frac{D}{u}\ seconds\]

Answer 2

so how do we work out the answer?

Answer 3

Yeah
area under speed time graph is distance
area of the triangle till t=20 is distance D
\[D=\frac{1}{2} \times 20 \times u\]
so
\[D=10u\]
so total time=\[20+\frac{D}{u}=20+10=30 seconds\]

Answer 4

I would do it in this way.
Distance travelled = area inder the graph
for the first 20s
D = (1/2) (20u) = 10u

2D = D + ut
20u = 10u + ut
10u = ut
t=10
So time needed = 10+20 = 30s

Answer 5

* under
+ I'm so lateeeee!!!!

Answer 6

i didnt get that part.
total time taken to cover 2D meters = 20 + D/u seconds

Answer 7

It's mentioned in the question that the car covers D meters in 20 seconds
After that it's speed is u m/s so to cover next D meters it'll take time
\[t=\frac{distance}{speed}=\frac{D}{u}\]
so total time to cover 2D meters
\[=20+\frac{D}{u}\]

Answer 8

okay why cant we write its as 2D/u + 20

Answer 9

Because the first D meters are covered at a constant acceleration, the car starts and accelerates till 20 seconds till u m/s and covers a distance D meters

Answer 10

@thushananth01 did you understand?

Answer 11

ok what about question b)

Answer 12

Yeah, Welcome back:)

Answer 13

?

Answer 14

Let's find the initial acceleration from t=0 to t=20 seconds
initial speed is =0 m/s
final speed is u m/s
time=20 seconds
so acceleration a
\[a=\frac{final-inital}{t}\]
\[a=\frac{u}{20}\]
Now it's given in second part at t=60 sec the car starts to slow down with an acceleration half as before so
here acceleration is actually deceleration , let it be a2
\[a2=-\frac{u}{40}\]
negative sign is there because there is deceleration so
initial velocity= u m/s
final velocity= u/4 m/s
we need to find how much time it takes to slow down from u to u/4
\[a=\frac{final-initial}{t}\]
Let's substitute the values
\[\frac{-u}{40}=\frac{u/4-u}{t}\]
we get
\[\frac{-u}{40}=\frac{-3u/4}{t}\]
or
\[t=40 \times \frac{3u}{4u}\]
we get
\[t=30 seconds\]
so it'll take t=30 seconds to slow down and at t=90 seconds (on graph) its speed will be u/4

Answer 15

a2=-u/40 because we half it right?

Answer 16

Yeah:)

Answer 17

"so it'll take t=30 seconds to slow down and at t=90 seconds (on graph) its speed will be u/4 "
cn u elaborate on this?

Answer 18

Sure
it'll decelerate from u m/s to u/4 at an acceleration of u/40 m/s^2 in 30 seconds

Answer 19

so then how did u get 90s cn u right equations?

Answer 20

at t=60 seconds its speed is u m/s and at t= (60+30) seconds its speed will be u/4
your answer is 30 seconds. On the graph at t=90 seconds its speed will be u/4

Answer 21

okai thnks i get it...awesome explanation

Answer 22

Welcome:D

Answer 23

can u help me with another doubt..

Answer 24

Yeah sure:)

Answer 25

@thushananth01 if you need to ask another question, close this one and post a new question:)

Answer 26

okai

Answer 27

:)

Answer 28

i ve posted :)