body starts moving from rest and moves with a uniform acceleration of 5 cm/s?. In how much time the velocity of body will become 50 cm/s and how much will it travel in this line?
Answers
Explanation:
Given:
\rightarrowtail \bold{Acceleration (a) = 5 cm/s}↣Acceleration(a)=5cm/s
\rightarrowtail \bold{Final\;velocity(v)=50cm/s}↣Finalvelocity(v)=50cm/s
\rightarrowtail \bold{Initial\;velocity(u)=0cm/s}↣Initialvelocity(u)=0cm/s
∵ Car is moving from rest.
\large\text{To find:}To find:
\rightarrowtail\bold{Time\;taken(t)}↣Timetaken(t)
\rightarrowtail \bold{Distance\;covered(S)}↣Distancecovered(S)
\large\text{Solution:}Solution:
Using the first equation of motion.
\mapsto\bold{v=u+at}↦v=u+at
Where we have.
\twoheadrightarrow\bold{v=\dfrac{50}{100}m/s }↠v=
100
50
m/s
\twoheadrightarrow\bold{a=\dfrac{5}{100}m/s^{2} }↠a=
100
5
m/s
2
\twoheadrightarrow\bold{u=0m/s}↠u=0m/s
Substituting we get.
\nrightarrow \bold{0.5 = 0 + 0.05(t)}↛0.5=0+0.05
\nrightarrow \bold{0.5=0.05t}↛0.5=0.05t
\nrightarrow\bold{t=\dfrac{0.5}{0.05} }↛t=
0.05
0.5
∴ Time taken (t) is 10 seconds.
Using the second equation of motion.
\mapsto \bold{S=ut+\dfrac{1}{2}at^{2} }↦S=ut+
2
1
at
2
Where we have.
\twoheadrightarrow \bold{u=0m/s}↠u=0m/s
\twoheadrightarrow \bold{a=0.05m/s^{2}}↠a=0.05m/s
- 2
\twoheadrightarrow\bold{t=10sec}↠t=10sec
Substituting we get.
\nrightarrow \bold{S=0(10)+\dfrac{1}{2}(0.05)(10)^{2} }↛S=0(10)+
2
1
(0.05)(10)
2
\nrightarrow \bold{S=0+0.025(100)}↛S=0+0.025(100)
∴ The distance covered is 2.5 meters.
Answer:
(i) t=10s (ii)s= 250cm
Explanation:
u=0 cm/s
v=50 cm/s
a=5 cm/s
t=?
w.k.t v = u+at ---> since acceleration is constant
so, 50= 0 + 5t
5t=50
t= 10s
for t=10s, distance covered will be;
s= ut + 1/2at^2
s= 0 + 1/2 × 5 × 100
s= 250 cm
Thus the body travels for 10s to attain a velocity of 50cm/s and travels 250cm during the 10s of motion
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