Question

A body which uniformly accelerated and have an initial velocity, travels 200 cm in the first 2 seconds and 220 cm in the next 5 seconds. Calculate the velocity of the body at the end of the 7th second from the start, assuming that the acceleration throughout the journey is uniform.

Kindly solve this question with clear an full explanation.​

Answer 1

Given :-

A body which is  uniformly accelerated and has an initial velocity travels 200 cm in the first 2 seconds and 220 cm in the next 5 seconds

To Find :-

Calculate the velocity of the body at the end of the 7th second from the start, assuming that the acceleration throughout the journey is uniform.

Solution :-

We know that

s = ut + 1/2 at²

In Case 1

200 = u(2) + 1/2 × a × (2)²

200 = 2u + 1/2 × a × 4

200 = 2u + 2a

Divide both sides by 2

200/2 = 2u + 2a/2

100 = u + a (i)

100 - a = u

In Case 2

Time = 2 + 5 = 7 sec

Total distance = 220 + 200 = 420 cm

420 = u(7) + 1/2 × a × (7)²

420 = 7u + 1/2 × a × 49

420 = 7u + 24.5a

420 = 7(100 - a) + 24.5a

420 = 700 - 7a + 24.5a

420 - 700 = -7a + 24.5a

-280 = 17.5a

-280/17.5 = a

-16 = a

Using 1

100 = u + a

100 = u + (-16)

100 = u - 16

100 + 16 = u

116 = u

Now

v = u + at

v = 116 + (-15)(7)

v = 116 - 105

v = 11 cm/s

Answer 2

$\huge \rm {Answer:-}$

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$\sf \red {Provided\: that:}$

★For the first two seconds(t=2seconds), distance travelled by the body (s)=200cm

★For the next five seconds(t=5 seconds), distance travelled by the body=220cm

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$\sf \blue {To\: Determine:}$

★The velocity of the body at the end of the 7th second

★Final velocity(v)=?

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★To calculate the final velocity we can the use the first kinematic equation

$\to \tt {\fbox{v=u+at}}$

★ But to use this equation,we need to first calculate acceleration(a) and intial Velocity (u)

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$\sf \pink {kinematic\: Equation\: to\: be\: used:}$

$\to \tt {s=ut+\frac{1}{2}at^{2}}$

★For the first 2 seconds,

Time (t)=2 seconds

Distance (s)=200cm

$\to \tt {200=u(2)+\frac{1}{2}a(2)^{2}}$

$\to \tt {200=2u+\frac{1}{2}4a}$

$\to \tt {200=2u+\frac{1}{\cancel{2}}\cancel{4}^{2}a}$

$\to \tt {200=2u+2a}$

$\to \tt {200=2(u+a)}$

$\to \tt {u+a=\frac{200}{2}}$

$\to \tt {u+a=100}$

$\to \tt {u=100-a}$----->Equation 1

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★For the next 5 seconds,

★Total Time =2seconds+5seconds=7seconds

★Total distance=200cm+220cm=420cm

$\to \tt {s=ut+\frac{1}{2}at^{2}}$

$\to \tt {420=u(7)+\frac{1}{2}a(7)^{2}}$

$\to \tt {420=7u+\frac{1}{2}49a}$

$\to \tt {420=7u+\frac{1}{\cancel{2}}\cancel{49}^{24.5}a}$

$\to \tt {420=7u+24.5a}$ ----->Equation 2

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★Substituting Equation-1 in Equation-2

$\to \tt {420=7(100-a)+24.5a}$

$\to \tt {420=700-7a+24.5a}$

$\to \tt {420-700=17.5a}$

$\to \tt {-280=17.5a}$

$\to \tt {a=\frac{-280}{17.5}}$

$\implies \tt \green {a=-16cm/s^{2}}$

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$\sf \orange {Now,:}$

★ Substituting the value of 'a' in Equation-1

$\to \tt {u=100-a}$

$\to \tt {u=100-(-16cm/s^{2})}$

$\to \tt {u=100+16cm/s^{2}}$

$\implies \tt {u=116cm/s}$

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★Intial Velocity (u) is determined to be 16cm/s

★Acceleration (a) is determined to be $\tt {-16cm/s^{2}}$

★ Time (t)=7 seconds

★using the first kinematic equation,

$\to \tt {v=u+at}$

$\to \tt {v=116+(-16)\times7}$

$\to \tt {v=116+(-112)}$

$\to \tt {v=116-112}$

$\implies \tt \green{\fbox{v=4cm/s}}$

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$\sf \purple {Therefore,}$

★The final velocity of the body at the end of the 7th second , assuming that the acceleration is uniform is 4cm/s.