Question

If the speed of a truck is reduced from 26.7m/s to 6.7m/s within a distance of 800m.Find how long were the brakes applied and how much longer would it take before coming to rest

Answer 1

Answer:

U=26.7m/s

V=6.7m/s

S=800m

Let the time be t

Answer 2

Brakes are applied for 47.96 seconds.

Explanation:

Initial speed, u = 26.7 m/s

Final sped, v = 6.7 m/s

Distance, d = 800 m

Firstly, lets find acceleration of the truck. Using third equation of motion as :

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(6.7)^2-(26.7)^2}{2\times 800}\\\\a=-0.417\ m/s^2$

Finding t using first equation of motion as :

$t=\dfrac{v-u}{a}\\\\t=\dfrac{6.7-26.7}{(-0.417)}\\\\t=47.96\ s$

Brakes are applied for 47.96 seconds.

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Kinematics

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