Question

an object moves 18m in the first 3s, 22m in the next 3s and 14m in the last 3s. what is its average speed​

Answer 1

6 m/s is the required answer.

CONSIDER:

• Let the distance be d
• Let the time be t

GIVEN:

• d1 = 18 m and t1 = 3 s (18m in the first 3s)

• d2 = 22m and t2 = 3 s (22m in the next 3s)

• d3 = 14 m and t3 = 3 s (14m in the last 3s)

TO FIND:

Average speed , v

FORMULA:

Average speed , v = total distance / total time

SOLUTION:

$v = \dfrac{d1 + d2 + d3}{t1 + t2 + t3} \\ \\ v = \dfrac{18 + 22 + 14}{3 + 3 + 3} \\ \\ v = \dfrac{54}{9} \\ \\ v = 6 \: \dfrac{m}{s}$

ANSWER:

Average speed is 6 m/s.

OTHER FORMULAE:

• Average velocity = ( final displacement - initial displacement) / time

• Average acceleration = (final velocity - initial velocity) / time

• Average speed = total distance / total time

• Differentiation of displacement gives instantaneous velocity

• Differentiation of velocity gives instantaneous acceleration.

• Integration of acceleration gives the change in velocity.

• Integration of velocity gives the displacement.