Question

a box slides down an incline with uniform acceleration. it starts from rest and attains a speed of 2.7 m/s in 3.0 s. find (a) the acceleration and (b) the distance moved in the first 6.0 s

Answer 1

Answer:

a.) 0.9m/s^2 b.) 16.2 m

Explanation:

Given,

u = 0m/s

v= 2.7m/s

t= 3 sec

Acceleration = v-u/t

$= \frac{2.7 - 0}{3 } \\ \frac{2.7}{3} = 0.9 \frac{m}{ {s}^{2} }$

(B) Time = 6sec

V = 2.7m/s

U = 0m/s

• distance = ut+1/2at^2

$= 0 \times 6 + \frac{1}{2} \times 0.9 \times {6}^{2} \\ = \frac{1}{2} \times \frac{9}{10} \times 36 \\ = 324 \div 20 = 16.2m$

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Answer 2

Given : u = 0m/s

             v= 2.7m/s

             t= 3 sec

To Find : a) the acceleration and (b) the distance moved in the first 6.0 s.

Solution :

(a) Here as ,

u = 0m/s

v= 2.7m/s

t= 3 sec

Acceleration a = $\[\frac{{v - u}}{t}\]$

acceleration = $\[\frac{{2.7 - 0}}{3}\]$ = 0.9 m/$\[{s^2}\]$

(b)

Distance moved when time = 6 sec

$\[s = ut + \frac{1}{2}a{t^2}\]$

putting all value in equation, when t = 6sec

s = 16.2 m.

Hence,  (a) the acceleration is 0.9 m/$\[{s^2}\]$ and (b) the distance moved in the first 6.0 s is 16.2 m.