First slide

Rectilinear Motion

Question

A particle moving along a straight line with a constant acceleration of 4 m/s² passes through a point A on the line with a velocity of +8 m/s at some moment. Find the
distance travelled by the particle in 5 seconds after that moment.

Moderate

A

26 m
B

18 m
C

9 m
D

32 m

Solution

$\begin{array}{l} u = + 8 m / s, a = - 4 m / s^{2} \\ v = 0 \\ \Rightarrow 0 = 8 - 4 t or t = 2 \sec \\ displacement in first 2 \sec . \end{array}$
$S_{1} = 8 \times 2 + \frac{1}{2} \cdot (- 4) \cdot 2^{2} = 8 m$
displacement in next 3 sec.
$S_{2} = 0 \times 3 + \frac{1}{2} (- 4) 3^{2} = - 18 m$
$distance travelled = |S_{1}| + |S_{2}| = 26 m$
ALITER: We can draw velocity-time graph of the situation.
Total distance is equal to magnitude of area under velocity-time graph.
$Hence, d = \frac{1}{2} \times 2 \times 8 + \frac{1}{2} \times 3 \times 12 = 8 + 18 = 26 m .$

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