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Question: Moving with uniform acceleration, a body covers \(150\,m\) during \(10\) seconds so that it covers \...

Moving with uniform acceleration, a body covers 150m150\,m during 1010 seconds so that it covers 24m24\,m during tenth second. Find initial velocity and acceleration of the body.
A. 2ms1;5ms22\,m{s^{ - 1}};5\,m{s^{ - 2}}
B. 5ms1;2ms25\,m{s^{ - 1}};2\,m{s^{ - 2}}
C. 3ms1;4ms23\,m{s^{ - 1}};4\,m{s^{ - 2}}
D. 4ms1;3ms24\,m{s^{ - 1}};3\,m{s^{ - 2}}

Explanation

Solution

To solve this question we should have a basic idea that it is about the motion of a body in a straight line. Hence we will solve this question using equations of motion. First we will find acceleration and then velocity. We will use the 2nd equation of motion to find a relation between acceleration and initial velocity and then using formula for distance covered in nth second we will find acceleration.

Formula used:
v=u+atv = u + at
where uu=initial velocity, aa=acceleration and vv=final velocity
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}
Where ss=distance travelled by the body, tt=time and uu=initial velocity.
sn=u+a2(2n1){s_n} = u + \dfrac{a}{2}\left( {2n - 1} \right)
Where sn{s_n} is distance travelled by a body in nth second, aa=acceleration, uu=initial velocity, n=nth second.

Complete step by step answer:
Let the body have initial velocity uu, acceleration aa, final velocity vv and may it cover ss distance and sn distance in time tt. Let us start solving this question by first finding relation between the acceleration and initial velocity of the body using the 2nd equation of motion:
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}
let us substitute the values from the question
150=u10+12a102150 = u10 + \dfrac{1}{2}a{10^2}
150=10u+12×a×100\Rightarrow 150 = 10u + \dfrac{1}{2} \times a \times 100
150=10u+50a\Rightarrow 150 = 10u + 50a
Divide the whole equation by 5.
30=2u+10a30 = 2u + 10a
3010a=2u..........(1)\Rightarrow 30 - 10a = 2u..........(1)

Now let us find acceleration using the formula for distance covered in 10th second,
sn=u+a2(2n1){s_n} = u + \dfrac{a}{2}\left( {2n - 1} \right)let us substitute the values;
24=u+a2(2×101)\Rightarrow 24 = u + \dfrac{a}{2}\left( {2 \times 10 - 1} \right)
24=u+a2×(19)\Rightarrow 24 = u + \dfrac{a}{2} \times \left( {19} \right)
Multiply both sides by 2:
48=2u+a(19)48 = 2u + a\left( {19} \right)
48=2u+19a\Rightarrow 48 = 2u + 19a................(2)
Now lets substitute the value of 2u from equation 1 into 2
48=3010a+19a48 = 30 - 10a + 19a
4830=19a10a\Rightarrow 48 - 30 = 19a - 10a
18=9a\Rightarrow 18 = 9a
a=2ms2\therefore a = 2\,m{s^{ - 2}}

Now let's find initial velocity using equation 1
3010a=u230 - 10a = u2
3010(2)=u2\Rightarrow 30 - 10\left( 2 \right) = u2
3020=u2\Rightarrow 30 - 20 = u2
2u=10\Rightarrow 2u = 10
u=102ms1\Rightarrow u = \dfrac{{10}}{2}\,m{s^{ - 1}}
u=5ms1\therefore u = 5\,m{s^{ - 1}}

Hence the correct answer is option B.

Note: Newton's equations of motion can be applied only when the body is moving with uniform acceleration. In the above questions the body is moving with uniform acceleration so we have applied Newton's equations. In some cases, when the acceleration is variable then we need to apply calculus for solving those problems.