Question

A Speed boat starts from rest and moves with constant acceleration of 5.0 m/seconds squared.. Find its speed and the distance traveled after 4 seconds has elapsed.

Answer 1

Answer :-

Speed => 20m/s

Distance travalled => 40m

Explanation :-

Given :

Initial velocity of the boat,u = 0 [as it starts from rest]

Acceleration,a = 5m/s^2

To Find :

Speed,v and distance,s after 4 seconds

Solution :

According to first law of motion,

$\sf{}v=u+at$

Put thier values and solve for “v”

$\sf{}\implies v=0+5\times 4$

$\sf{}\therefore v=20m/s$

Therefore,speed is equal to 20m/s

According to second law of motion,

$\sf{}s=ut+\dfrac{1}{2}at^2$

Put thier values and solve for “s”

$\sf{}\implies s=0\times 4+\dfrac{1}{2}\times 5\times 4^2$

$\sf{}\implies s=0+\dfrac{1}{2}\times 5\times 16$

$\sf{}\implies s=0.5 \times 5\times 16$

$\sf{}\therefore s=40m$

Therefore,distance travelled is equal to 40m

Answer 2

Given:

• u= 0m/s (as it starts from rest)
• a= 5$m/s^2$
• t= 4s

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Need to find:

• v=?
• s=?

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Solution:

From first equation of motion, We know,

$v = u + at$

$\implies \: v = 0 + 5 \times 4$

$\red{\bold{ \implies \: v = 20 m/s}}$

━━━━━━━━━━━━━━━━━━

From 2nd equation of motion, We know,

$s = ut + \dfrac{1}{2} \times a {t}^{2}$

$\implies \: s = 0 \times 4 + \dfrac{1}{2} \times 5 \times {4}^{2}$

$\implies \: s = \dfrac{1}{2} \times 5 \times 16$

$\red{\bold{ \implies \: s = 40m}}$