Question

Acceleration-time graph of a particle moving in a straight line is as shown in figure. At time $t=0$, velocity of the particle is zero. Find the average acceleration in a time interval from $t=6$s to $t=12$s.

• A. $5\frac{m}{s^2}$
• B. $20\frac{m}{s^2}$
• C. $10\frac{m}{s^2}$
• D. $50\frac{m}{s^2}$

Answer 1

The correct option is A $5\frac{m}{s^2}$

We know that velocity is the area under the acceleration- time graph.

At $t=6s-$

$v_1=10\times 4+\frac{1}{2}(10+20)\times 2=70m/s$

At $t=12s$

$v_2=10\times 4+\frac{1}{2}(10+20)\times 2+\frac{1}{2}\times 4\times 20-\frac{1}{2}\times 10\times 2=100m/s$

Time interval$=\Delta t=12-6=6s$

Average acceleration$=a_{av}=\frac{v_2-v_1}{\Delta t}$

$⟹a_{av}=5m/s^2$