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Solution:- Given in the question that
x=18t+5t2
(i) we know that velocity
v=dxdt=ddt(18t+5t2)=18+10t
To find velocity at t=2s, put t=2 in above equation.
So, Instantaneous velocity=18+10×2=38m.s2
(ii) displacement at t=2 is
x1=18×2+5×22=56m
At t=3
x2=18×3+5×33=54+45=99m
average velocity
v=x2−x1t2−t1=99−563−2=43ms−1
(iii)
a=dvdt=ddt(18+10t)=10ms−2.
Solution :-
(i) let the car accelerate for the time t, and v be the maximum velocity of the car at time t1.
We know that
v=u+at
Therefore, v=0+αt1
or,
t1=vα(1)
Now starting with maximum velocity v, the car decelerates at constant rate β and comes to rest in time (t−t1). Therefore,
0=v−β(t−t1)
or,
t−t1=vβ(2)
Adding equations (1) and (2)
t=vα+vβ=v(α+βαβ)
or,
v=αβtα+β(3)
This gives maximum velocity attained by the car.
(ii) distance covered by the car in time t1 is
x1=0+12αt21
By using equation of motion
x=ut+12at2
x1=12α⋅v2=−v22α
[using 1]
Distance travelled by the car in time (t−t1)
x=v(t−t1)−12β(t−t1)2=v⋅vβ−12βv2β2=v22β (using epn2)
Therefore total distance travelled by the car is
x=x1+x2=v22α+v22β=v22[1α+1β]
Using equation (3) and solving for x we get
x=αβt22(α+β)
Solution:-
Here u=0,t=25s,
Given that
v=180Km/h=180×518=50m/s
Acceleration
a=v−ut=50−025=2ms2
Distance,
s=ut+12at2=0×25+12×2×(25)2=625m.
Solution :-
(i) Here it is given that
s=2m,t=4s,u=0
as
s=ut+12at2
Putting respective values we get
2=0+12×a×42
or,
a=0.25m/s2
(ii) Time taken to cover the first metre of track is given by
1=0+12×0.25×t2
⇒t2=8
Or t=2√2=2×1.414=2.83s
Hence time taken to cover the second meter of track is
=4−2.83=1.17s.
(iii) Speed at the bottom,
v=u+at=0+0.25×4=1m/s
Question 1- What do you understand by term acceleration and retardation distinguish between average acceleration and instantaneous acceleration.
Question 2- Represent graphically and explain the motion of an object when the object is under the following conditions
(i) object is at rest
(ii) object with uniform motion along straight line
(iii) object with accelerated motion along straight line
(iv) object with decelerated motion moving along a straight line.
Choose correct option to answer following questions.
Question 1 :- A body is covering distance in proportion to square of time. The acceleration of the body is
(a) increasing
(b) decreasing
(c) zero
(d) constant
Answer:- (d)
Question 2:- The relation between t and distance x is t=αx2+βx. Where α and β are constants.
The retardation is
(a) 2αv3
(b) 2αβv3
(c) 2βv3
(d) 2β2v3
Answer :- (a)
Given that
t=αx2+βxdtdt=2αxdxdt+βdxdt1=2αxv+βvv⋅(2αx+β)=1(2αx+β)=1v
Again differentiating both the sides w.r.t. t we get
2αdxdt=−v−2⋅dvdt2αv=−v−2⋅acceleration
⇒acceleration=−2αv3
Hence,
retardation=+2αv3
Solution : - (a)
From the given velocity-displacement graph,
slope=v0x0 ,
and intercept on y-axis isv0
Thus equation for this graph will be
v=−v0x0x+v0[y=mx+c]dvdt=−v0x0dxdta=−v0x0.v=−v0x0[−v0x0x+v0]=v20x20x−v20x0
From this we can clearly see that graph must have positive slope (v02x2) and negative intercept (−v20x0) on y-axis.
Solution :- (c)
For accelerated motion
u=0,a=fands=s.
As,
v2−u2=2as
∴ v_{1}^{2}-0^2=2fs
or,
v_1=\sqrt{2fs}
For uniform motion
u=v_1=\sqrt{2fs}, \, t=t
Distance travelled,
s_2=ut=\left( \sqrt{2fs} \right) t
For decelerated travelled
u=\sqrt{2fs}, \, a=-f/2, \, v=0
As, v^2-u^2=2as
0^2-\left( \sqrt{2fs} \right) ^2=2\times \left( -f/2 \right) s_3
distance travelled,
s_3=2s
given in the question total distance travelled is
s+s_2+s_3=5s
s+\left( \sqrt{2fs} \right) t+2s=5s
or,
\sqrt{2fs}t=2s
Squaring both sides me get
s=\frac{1}{2}ft^2
Answer :- (c)
\begin{align*}
a=\frac{dv}{dt}=bt
\\
\therefore v=\int{\sqrt{\left( bt \right)}dt}=\frac{bt^2}{2}+c
\end{align*}
Initially at t=0, \, v=v_0
\Rightarrow c=v_0
Hence,
v=\frac{dx}{dt}=\frac{bt^2}{2}+v_0
Or,
\begin{align*}
x&=\int{\left( \frac{bt^2}{2}+v_0 \right)}dt
\\
&=\frac{bt^3}{6}+v_0t+C'
\end{align*}
At t=0,\, x=0
\therefore C'=0
Therefore,
x=v_0t+\frac{1}{6}bt^3
Question 1 :- Give an example which shows that a positive acceleration can be associated with a slowing down object.
Question 2 :- is the acceleration of a car greater than when accelerator is pushed to the floor or when break pedal is pushed hard.
Question 3 :- suppose the acceleration of a body varies with time. Then what does area under its acceleration – time graph for any time interval represent.
Question 4 :- The ϑ –t graphs of two objects make angle of and with the time axis. Find the ratio of their acceleration.
Question 5 :- is it possible that your cycle has northward velocity but southward acceleration? If yes, how?