12 questions · self-marked practice — reveal the answer and mark yourself.
$=4(5)+\frac{1}{2}(1.2)5^2=35\text{m}.$
$\text{V}_{\text{ave}}=\frac{\text{displacement}}{\text{times}}=\Big(\frac{\overrightarrow{\text{AB}}}{\text{t}}\Big)$
t = time We can see that $\overrightarrow{\text{AB}}$ is along $\overrightarrow{\text{BC}}$ i.e. they are in same direction. The point is B(5m, 3m).
$\therefore\text{S = ut}+\frac{1}{2}\text{at}^2$
$\Rightarrow12.1=0+\frac{1}{2}(9.8)\times\text{t}^2$
$\Rightarrow\text{t}^2=\frac{12.1}{4.9}=2.46$
$\Rightarrow\text{t}=1.57\text{sec}$
For cadet velocity = 6km/hr = 1.66m/sec Distance = vt = 1.57 × 1.66 = 2.6m. The cadet, 2.6m away from tree will receive the berry on his uniform.
time = 10 sec.
$\text{V}_{\text{ave}}=\frac{\text{s}}{\text{t}}=\frac{100}{10}=10\text{m/s}.$
at 2 sec. Vinst = 20m/s.
At 5 sec it is at rest.
Vinst = zero.
At 8 sec it is moving with uniform velocity 20m/s
Vinst = 20m/s
At 12 sec velocity is negative as it move towards initial position. Vinst = -20m/s.
