Question

answer this question​

Answer 1

(1) what is the distance travelled by the object in the first two seconds?

In velocity - time graph,

Distance covered by the object

= area under the v - t graph

Distance travelled in 2 sec

= Area of ∆ ABE

= 1/2 × base × height

= 1/2 × AE × BE

= 1/2× 2 × 15

= 30/2

= 15 m

Distance travelled in 2 sec = 15 m

(2) calculate the total distance travelled by the object?

Total distance

= Area of ∆ ABE + Area of EBCF + Area of triangle ∆ FCD

= 1/2 × AE× BE + EB × BF + 1/2 × FD× CF

= 1/2× 2 × 15 + (5-2)×15 +1/2× (6-5) × 15

= 15 + 15 ×3 + 15/2

= 15 + 45+ 7.5

= 67.5 m

Total distance travelled by the object= 67.5 m

(3) find the acceleration between A and B B and C, C and D

the slope of the V-T graph = acceleration

A and B

Slope = tan x = BE / AE =15/2= 7.5 m/s ²

Slope = acceleration = 7.5 m /s ²

B and C

Slope = 0

acceleration = 0

A and B

Slope = tan x = CF/ FD =15/1= 15 m/s ²

Slope = acceleration = 15 m /s ²

Answer 2

HEY MATE HERE IS YOUR ANSWER.

ACCORDING TO THE QUESTION

1) Distance travelled in first two seconds= Area of ∆ABE

S = ½ × base × height

S = ½ × 2 × 15

S = 15m

2) Total distance = Area of ∆ABE+Area of BCEF + Area of ∆CFD

S = 15 + 15×3 + ½×15×1

S = 15 + 45 + 7.5

S = 67.5m

3) Acceleration from A to B = perpendicular/base

= 15/2

= 7.5m/s^2

Acceleration from B to C = it is constant so acceleration will be 0m/s^2

Acceleration from C to D = 15/1

= 15m/s^2

Hope this will help you.

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