A car starts with velocity of 3m/s and moves in a straight line with a constant acceleration if the velocity at the end of 5sec is 5.5 m/s find the uniform acceleration and the distance travelled in 10sec

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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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