Question

A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec

• A. 8.3 m
• B. 9.3 m
• C. 10.3 m
• D. None of above

Answer 1

Answer: (A)

8.3 m

Answer 2

Let initial $(t = 0)$ velocity of particle = $u$

for first 5 sec of motion $s_{5} = 10metre$,

so by using $s = ut + \frac{1}{2}at^{2}$

$10 = 5u + \frac{1}{2}a(5)^{2}$⇒ $2u + 5a = 4$ …. (i)

for first 8 sec of motion $s_{8} = 20metre$

$20 = 8u + \frac{1}{2}a(8)^{2}$ ⇒ $2u + 8a = 5$ .... (ii)

By solving (i) and (ii) $u = \frac{7}{6}m/sa = \frac{1}{3}m/s^{2}$

Now distance travelled by particle in total 10 sec.

$s_{10} = u \times 10 + \frac{1}{2}a(10)^{2}$

by substituting the value of $u$ and $a$ we will get

$s_{10} = 28.3$m

So the distance in last 2 sec = s₁₀ – s₈ $= 28.3 - 20 = 8.3$m