Question

study the velocity time graph and calculate
a.the acceleration from a to b
b.the acceleration from b to c.
c.the distance covered by region ABE and BCFE
d.average velocity from c to d

Answer 1

(a) a = (25 - 0) / (3 - 0) = 8.3 m/s2

(b) a = (20 - 25 ) / (4 - 3) = -5 m/s2

(c) Distance = Area of triangle ABE = 1 / 2 × 3 × 25 = 37.5 m 

(d) V = (20 - 0) / 2 = 10 m/s

(e) This Distance = Area of trapezium BCFE 

= 1 / 2 (25 + 20) × (4 - 3) = 22.5 m 






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Answer 2

Answer:

The given diagram depicts a motion in which a particle starts from point A at time t=0 and then it accelerates till it reaches point B. From point B the particle decelerates till t=6s . On the X-axis we have time and, on the Y-axis, we have velocity.

we have the kinematical equation,

v−u=at

v is the final velocity,

u is the initial velocity,

t is the time taken and

a is the acceleration.

1.From the graph we have the following values when the particle travels from A to B.

v=25ms−1 , u=0ms−1 and t=3s ,

substituting these values in the above equation, we have

$a= \frac{25−0}{3}$

$a= \frac{25}{3} =8.33 \: ms−2$

Therefore, the acceleration from A to B is 8.33ms−2 .

2.The acceleration from B to C.

2.The acceleration from B to C.From the graph we have the following values when the particle travels from B to C.

v=20ms−1 ,

u=25ms−1 and

t=1s ,

substituting these values in the above equation, we have

$a= \frac{20−25}{1}$

⇒a=−5ms−2

The acceleration from B to C is −5ms−2 .

The acceleration from B to C is −5ms−2 .The negative sign implies that the body is decelerating.

3.Distance covered in region ABE.

Distance covered in region ABE.Distance covered is given as the area under the velocity time graph.

Distance covered in region ABE.Distance covered is given as the area under the velocity time graph.Distance covered s will be given as

Distance covered in region ABE.Distance covered is given as the area under the velocity time graph.Distance covered s will be given ass=Area(△ABE)

$⇒s= \frac{1}{2} ×3×25$

⇒s=37.5m

The distance travelled in the region ABE is 37.5m .

4. The average velocity from C to D.

The average velocity from C to D.The average velocity will be

$v_avg= \frac{20}{2}$

$v_avg= \: 10ms−1$

The average velocity from C to D is 10ms−1 .

5.Distance covered in the region BCFE

Distance covered in the region BCFEDistance will be given as the area under BCFE region, the BCFE region comprises of triangle with height 5m ,

base EF=1m and

a rectangle have having length

CF=20m and breadth EF=1m

The distance covered s will be equal to the total area, it will be given as

$s= \frac{1}{2} ×1×5+20×1$

⇒s=22.5m

⇒s=22.5mThe distance covered in the region BCFE is 22.5m

Thus,

a) the acceleration from A to B is 8.33ms−2

b) acceleration from B to C is −5ms−2 .

c) The distance travelled in the region ABE is 37.5m .

d) The average velocity from C to D is 10ms−1 .

e) The distance covered in the region BCFE is 22.5m.

Project code #SPJ2