Question:
If the displacement of a body moving with constant acceleration in a straight line path in the first four seconds of time is 56 m and its displacement in the fourth second of the motion is 17 m, then the average velocity of the body during sixth second of its motion is
If the displacement of a body moving with constant acceleration in a straight line path in the first four seconds of time is 56 m and its displacement in the fourth second of the motion is 17 m, then the average velocity of the body during sixth second of its motion is
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Displacement during nth second:
\[
S_n=u+\frac{a}{2}(2n-1)
\]
is very useful in such problems.
Updated On: Jun 17, 2026
- $18\,ms^{-1}$
- $21\,ms^{-1}$
- $11\,ms^{-1}$
- $26\,ms^{-1}$
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The Correct Option is D
Solution and Explanation
Concept:
For uniformly accelerated motion,
\[
S_n=u+\frac{a}{2}(2n-1)
\]
gives displacement during the nth second.
Step 1: Use displacement during fourth second.
\[ 17=u+\frac{a}{2}(7) \] \[ 17=u+3.5a \] \[ u=17-3.5a \]
Step 2: Use displacement in first four seconds.
\[ 56=4u+\frac12 a(4)^2 \] \[ 56=4u+8a \] Substituting \(u=17-3.5a\), \[ 56=4(17-3.5a)+8a \] \[ 56=68-14a+8a \] \[ 6a=12 \] \[ a=2\,ms^{-2} \] Hence, \[ u=17-7=10\,ms^{-1} \]
Step 3: Find displacement during sixth second.
\[ S_6=u+\frac{a}{2}(11) \] \[ S_6=10+11 \] \[ S_6=21\,m \] Since sixth second lasts for one second, \[ \text{Average velocity}=21\,ms^{-1} \] \[ \boxed{21\,ms^{-1}} \]
Step 1: Use displacement during fourth second.
\[ 17=u+\frac{a}{2}(7) \] \[ 17=u+3.5a \] \[ u=17-3.5a \]
Step 2: Use displacement in first four seconds.
\[ 56=4u+\frac12 a(4)^2 \] \[ 56=4u+8a \] Substituting \(u=17-3.5a\), \[ 56=4(17-3.5a)+8a \] \[ 56=68-14a+8a \] \[ 6a=12 \] \[ a=2\,ms^{-2} \] Hence, \[ u=17-7=10\,ms^{-1} \]
Step 3: Find displacement during sixth second.
\[ S_6=u+\frac{a}{2}(11) \] \[ S_6=10+11 \] \[ S_6=21\,m \] Since sixth second lasts for one second, \[ \text{Average velocity}=21\,ms^{-1} \] \[ \boxed{21\,ms^{-1}} \]
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