Question

A body travels 200 cm in the first two seconds and 220 cm in the next 4 seconds with same acceleration. The velocity of the body at the end of the 7th second is:

• A. 5 cm/s
• B. 10 cm/s
• C. 15 cm/s
• D. 20 cm/s

Answer 1

The correct option is B 10 cm/s

let the body have displacement as follows :

$x=a{t}^2+bt+c$ (a = constant acceleration)

as body travels $200cm$ in first $t=2$ seconds,

$\Rightarrow x_2-x_1=(a{t}_2^2+b{t}_2+c)-(a{t}_1^2+b{t}_1+c)$

$\Rightarrow 200=4a+2b\to \left(1\right)$

Similarity it travels $220cm$ in $t_1=25$ to $t_2=65$

$\Rightarrow 220=(a{6}^2+b6+c)-(a{.2}^2+b.2+c)$

$\Rightarrow 220=32a+4b\to \left(2\right)$

solving $\left(1\right)$ & $\left(2\right)$ given,

$a=\frac{-15}{2}$ and $b=115$

Thus, equation of motion for particle,

$x\left(t\right)=-\frac{15}{2}{t}^2+115t+c$

and, velocity at $t=7$ given by:

$v(t=7)=\frac{dx}{dt}{|}_{t=7}=(-15t+115{)}_{t=7}=10cm/s$