Question

please solve it................​

Answer 1

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Maximum velocity of the object will be At m/sec

Explanation:

We have that firstly the object starts from rest so its initial velocity will u = 0 m /sec

Its acceleration is Am/sec^2 and deceleration is Bm/sec^2'

As it is given that from rest its first accelerate and then decelerate so just before deceleration object will be at maximum speed

From first equation of motion we know that v = u+at

So v=0+At=Atm/sec

So maximum velocity of the object will be At m/sec

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Explanation:

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

Distance traveled by the body in the first second=

                                                                                 =5 meters= S1

Distance traveled by the body in the second second= distance traveled by the body in two seconds- distance traveled by the body in one second

= $</p><p>0 \times 2 + \frac{1}{2} \times 10 \times {2}^{2} - 5$

=20-5

=15 meters= S2

Distance traveled by the body in the third second= distance traveled by the body in three seconds- distance traveled by the body in two seconds

= $0 \times 3 + \frac{1}{2} \times 10 \times 3 ^{2} - 20$

=45-20

=25 meters=S3

So, answer is.

S1 : S2 : S3 = 5 : 15 : 25

=> 1 : 3 : 5

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