Question

In figure shown, the graph shows the variation of a unidirectional force Facting on a body of mass 10 kg
(in gravity free space), with timet. The velocity of the body att=0 is zero. (Area under F-tourve gives
change in momentum)
1) The velocity of the body at t=30 s is
(A) 30 m/s
(B)20 m/s
(C) 40 m/s L (D) none

2) the power of the force at t=12 is (power=Force ×velocity)?

3) the average acceleration of the body from t=0 to t=15 is?
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Answer 1

Sorry i can only tell 1st part

Area under curve gives change in velocity

i.e, v - u

area = 1/2*2*10 + 1/2*1*5 + 1*5 + 10 + 1/2*5*1

= 10 + 2.5 + 5 + 10 + 2.5

= 30

velocity at t = 30s is 30m/s

Answer 2

The velocity of the body at t = 30 s is (B) 30 m/s.

The power of the force at t = 12 s is 217.6 W.

The average acceleration will be 0.5 m/s².

Explanation 1

We have been given a Force- time graph . According to Newton's second law of motion, we know that

$F=\frac{dP}{dt}$    $;$ $or$

$dP = F$ × $dt$

We know, mass is a common factor in both force and momentum hence,

If we solve further, we get

$m.dv=F.dt$   $;$ $or$

$dv=\frac{F.dt}{m}$

Hence, we understand that the product of force and time divided by mass of the object gives us the velocity.

Therefore, velocity at $t=30s$ will be sum of all areas from $t=0s$ to $t=30s$.

$=\frac{\frac{1}{2}(10)(20)+\frac{1}{2}(5)(10)+(5)*(10)+(10)*(10)+\frac{1}{2}(5)(10)}{10}$

$=\frac{300}{10}$

$=30m/s$

Explanation 2:

We know, at relation between power and force is $P=F$ × $v$

We can see that the value of force at $t=12s$ is $16N.$

Now, velocity at $t=12s$ is

$=\frac{\frac{1}{2}(10)(20)+\frac{1}{2}(2)(4)+(2)*(16) }{10}$

$=\frac{136}{10} =13.6m/s$

Hence, power $P=$ $(16)$ × $(13.6)$ $=217.6W$

Explanation 3

We know, the differential form of acceleration is given by $a=\frac{dv}{dt}$.

Velocity at $0s$ = 0 m/s

Velocity at $10s$ = $\frac{\frac{1}{2}(10)(20) }{10} = 10m/s$

Velocity at $15s$ = $\frac{\frac{1}{2}(5)(10)+(5)*(10) }{10} =7.5m/s$

Hence,

Acceleration between $t=0s$ to $t=10s$ ,

$a_{1}=\frac{10-0}{10}=1m/s^{2}$

Similarly, acceleration between $t=10s$ to $t=15s$

$a_{2}=\frac{7.5-10}{5} =-0.5m/s^{2}$

Hence, average acceleration will be $0.5m/s^{2}$.

Although your question is incomplete, you might be referring to the diagram below.