First slide

Rectilinear Motion

Question

A body moving with a uniform acceleration crosses a distance of 15m in the 3rd second and 23m in the 5th second. The displacement in 10sec will be

Easy

A

150m
B

200m
C

250m
D

300m

Solution

$g i v e n S_{3} = 15 m; l e t i n i t i a l v e l o c i t y = u; a c c e l e r a t i o n = a; s u b s t i t u t e i n S_{n} = u + \frac{a}{2} (2 n - 1) 15 = U + \frac{1}{2} a (2 \times 3 - 1) = U + \frac{5 a}{2} \to (1) g i v e n S_{5} = 23 m; s u b s t i t u t e i n S_{n} = u + \frac{a}{2} (2 n - 1)$
$23 = U + \frac{1}{2} a (2 \times 5 - 1) = U + \frac{9 a}{2} \to (2)$
Solving (1) and (2)
a = 4m/ $s^{2}$ , U = 5m/s
Displacement in 10 sec
$S = U t + \frac{1}{2} a t^{2}$
$= 5 \times 10 + \frac{1}{2} \times 4 \times 10^{2}$
S= 250 m=Displacement in 10 sec

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