Question

The velocity-time graph of an object moving along
straight line is as shown
Velocity (m/sec)
Time (c)
Calculate distance covered by object between
(a) - 0 - 5 sec. (b) 0 10 se​

Answer 1

Answer:

Given:

A velocity-time graph has been provided.

To find:

Distance travelled in

• from 0 - 5 secs
• from 0 - 10 secs

Concept:

Always remember that the area under the velocity - time curve gives us distance.

Calculation:

From 0 - 5 seconds:

Distance travelled

= Area of triangle + area of rectangle

= (½ × 2 × 20) + (3 × 20)

= 20 + 60

= 80 metres

From 0 - 10 seconds:

Distance travelled

= Area of triangle + area of rectangle + area of triangle

= (½ × 2 × 20) + (3 × 20) + (½ × 5 × 20)

= 20 + 60 + 50

= 130 metres.

Answer 2

$\huge{\orange{\underline{\red{\mathbb{FIND:-}}}}}$

DISTANCE TRAVELLED,

0-5S.

0-10S.

$\huge{\orange{\underline{\red{\mathbb{ANSWER:-}}}}}$

CASE. 1)

Q. DISTANCE TRAVELLED BY 0-5s.

ACCORDING TO DIAGRAM :-

$\bold\rightarrow$ triangle area+reactangle area

$\bold\rightarrow$(½×b×h)+(l×b)

(put values)

$\bold\rightarrow$(½×2×20)+(3×20)

$\bold\rightarrow$ 20+60

$\huge \bold\red\rightarrow$ 80m.

CASE. 2)

Q. DISTANCE TRAVELLED BY 0-10s.

ACCORDING TO DIAGRAM :-

$\bold\rightarrow$ triangle area +reactangle area+triangle area

$\bold\rightarrow$ $\bold\rightarrow$(½×b×h)+(l×b)+(½×b×h)

(put values)

$\bold\rightarrow$ (½×2×20)+(3×20)+(½×5×20)

$\bold\rightarrow$20+50+60

$\huge\bold\red\rightarrow$130m.

$\huge{\orange{\underline{\red{\mathbb{THANKS.}}}}}$