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# A body covers a distance of 4 m in 3rd second and 12 m in 5th second. If the motion is uniformly accelerated, how far will it travel in the next 3 seconds?(A) 10 m(B) 30 m(C) 40 m(D) 60 m

Last updated date: 20th Jun 2024
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Hint:Use the formula to determine the distance of the particle in nth second as determine the initial velocity and acceleration of the particle. Use the kinematic relation to determine the distance travelled by the particle in the next 3 second using quantities you obtained above.

Formula used:
$\Rightarrow {S_n} = u + \dfrac{1}{2}a\left( {2n - 1} \right)$
Here, u is the initial velocity of the body, a is the uniform acceleration and n is the value of second for which the distance is to be calculated.

We know that the displacement of the uniformly accelerated body is given by the relation,
$\Rightarrow {S_n} = u + \dfrac{1}{2}a\left( {2n - 1} \right)$
Here, u is the initial velocity of the body and a is the uniform acceleration.
We have given, the distance is 4 m in the 3rd second. Therefore, we can write,’
$\Rightarrow 4\, = u + \dfrac{1}{2}a\left( {2\left( 3 \right) - 1} \right)$
$\Rightarrow 4 = u + \dfrac{5}{2}a$ …… (1)
Also, the distance is 12 m in the 5th second. Therefore, we can write,
$\Rightarrow 12 = u + \dfrac{1}{2}a\left( {2\left( 5 \right) - 1} \right)$
$\Rightarrow 12 = u + \dfrac{9}{2}a$ …… (2)
Subtract equation (1) from equation (2).
$\Rightarrow 12 - 4 = u + \dfrac{9}{2}a - \left( {u + \dfrac{5}{2}a} \right)$
$\Rightarrow 8 = 2a$
$\Rightarrow \therefore a = 4\,m/{s^2}$
Solve the simultaneous equations (1) and (2) from u, we get,’
$\Rightarrow u = - 6\,m/s$
Now, determine the final velocity of the body at the end of 5 second as follows,
$\Rightarrow v = u + at$
Substitute $u = - 6\,m/s$, $a = 4\,m/{s^2}$ and $t = 5\,s$ in the above equation.
$\Rightarrow v = - 6\,m/s + \left( {4\,m/{s^2}} \right)\left( {5\,s} \right)$
$\Rightarrow v = 14\,m/s$
The distance travelled by the body in the next 3 second can be calculated using the kinematic relation,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
The final velocity at the end of 5 second is the initial velocity for this case.
Substitute $v = 14\,m/s$, $t = 3\,s$, and $a = 4\,m/{s^2}$ in the above equation.
$\Rightarrow s = \left( {14\,m/s} \right)\left( {3\,s} \right) + \dfrac{1}{2}\left( {4\,m/{s^2}} \right){\left( {3\,s} \right)^2}$
$\Rightarrow s = 42\,m + 18\,m$
$\Rightarrow \therefore s = 60\,m$

So, the correct answer is option (D).

Note: Since the acceleration is uniform, the acceleration is the same at any position. The final velocity at a given second is also the initial velocity of the particle at that second due to uniform acceleration. The formula for determining the distance of the particle at nth second is obtained from the relation $s = ut + \dfrac{1}{2}a{t^2}$.