Answer:
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.........Answer.........
Given:-
u = 0.3 m/s
v = 5.5 m/s
t = 5 seconds
Explanation:-
Case-1
Acceleration = ?
From the first law of motion.
v = u + atv=u+at
a = \frac{v - u}{t}a=
t
v−u
a = \frac{5.5 - 0.3}{5}a=
5
5.5−0.3
a = \frac{5.2}{5}a=
5
5.2
a = 1.04a=1.04
\boxed{\boxed{a = 1.04 m/s^2}}
a=1.04m/s
2
So,the acceleration produced is 1.04 m/s^2.
Case-2
Distance travelled = ?.
From second equation of motion.
s = ut + \frac{1}{2} a {t}^{2}s=ut+
2
1
at
2
s = 0.3 \times 10 + \frac{1}{2} \times1.04 \times 100s=0.3×10+
2
1
×1.04×100
s = 3 + \frac{1}{2} \times 104s=3+
2
1
×104
s = 3 + 52s=3+52
s = 55 \: meterss=55meters
\boxed{\boxed{s = 55 meters}}
s=55meters
So,the distance travelled is 55 meters.
Explanation:
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