Question

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Answer 1

Answer:

Explanation:

69. Here, first the acceleration is given, so, we must find the area of the triangle, in the graph with the corner points as 10, 11

Area of triangle= 1/2 x base x height

so, the answer will be 1/2 x 10 x 11= 55 m/s

70. To find distance covered, again calculate the area of this graph, which is equal to the area of the spaces enclosed by the x and y axes. So distance covered = (4x2) + (2x2) + (2x2) =16m

In order to find displacement, calculate area of spaces with positive velocity and subtract area enclosed with negative velocity, thus it will give you the answer as:

(8 + 4) - (2x2) = 8m.

71. distance covered in last two sec =triangle CD5

=1/2 x base x height

=1/2 x 2 x 8

=8m

now, total distance covered in seventy seconds

=area of trapezium

=1/2 (sum of parallel sides) x height of the trapezium

=1/2(2+6)×8

=1/2×8×8

=32 m

Now, the fraction of distance covered in last two seconds and the  total distance is:

$\frac{8}{32}$ = $\frac{1}{4}$  

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Answer 2

Answer:

69. u= 0

a(maximum) = 10 m/s²

t = 11 s

v= u+at

v= 0+(10×11)

v= 110m/s

70. area of v-t graph is distance /displacement

for 0to 2s

= 4×2 = 8m

2to 4s

= 2×2

= 4m

(0to 4s, distance = 8+4 = 12m , displacement

= 8-4 = 4m as it is in opposite direction)

4to 6s

2×2

= 4m

distance = 12+4= 16m

displacement = 4+4= 8m

71. last two seconds

= area of triangle

= 1/2×10×2

= 10m

total distance=

area of trapezium

=1/2×(3+7)×10

1/2×10×10

= 9×5

= 45

fraction=

10/45 = 2/9