Question

A body starts from rest with an acceleration 2m/s^2 till it attains the maximum velocity then retards to rest with 3m/s^2.if total time taken is 10 seconds then maximum speed attained is

Answer 1

U1 = 0
a1 = 2m/s²
v1 = ?
t1 = t

V1 = U1 + a1t
V1 = at
V1 = 2t1

when it just begun to start regarding it's speed then ,
U2 = V1 = 2t1
V2 = 0 m/s ( because finally it is at rest)
t2 = t2
a = -3m/s²

V2 = U2 + a2t2
0 = 2t1 + (-3)(t2)
-2t1 = -3t2
2t1 = 3t2
t1/t2 = 3/2
t1 : t2 = 3:2
t1 = 3
t2 = 2

but t1+t2 = 10s
hence ,
t1 = 3×2 = 6
t2 = 2×2 = 4

Vmax = V1 = 2t1 = 2(6) = 12m/s
_________________
a = acceleration
t = time
U = initial speed
V = final speed

Answer 2

$\boxed{\mathsf{ Solution : }}$

Let it accelerate for t seconds.

So, it retards for ( 10 - t ) second.

Now, when it starts to accelerate.

⇒ Initial Velocity ( u ) = 0

⇒ Acceleration ( a ) = 2 m/s²

⇒ Time ( t ) = t sec

⇒ Final Velocity / Maximum Velocity ( v ) = ?

Using 1st equation of Motion ,

⇒ v = u + at

⇒ v = 0 + ( 2m/s² ) × t

$\therefore$ v = ( 2mt /s² )

When it starts retarding,

⇒ Initial Velocity ( $\mathsf{u_1 }$ ) = ( 2mt/s² )

⇒ Final Velocity ( $\mathsf{v_1}$ ) = 0

⇒ Acceleration ( $\mathsf{a_1}$ ) = -3 m/s²

⇒ Time ( $\mathsf{t_1}$) = ( 10 - t ) sec

Using 1st equation of Motion,

⇒ v = u + at

⇒ 0 = ( 2 mt/s² ) + ( -3 m/s² ) ( 10 - t )

⇒ 0 = ( 2 mt/s² ) - ( 30 m/s² ) + ( 3 mt/s² )

⇒ 0 = ( 5 mt/s² ) - ( 30 m/s² )

⇒ 5 mt/s² = 30 m/s²

⇒ 5t = 30

⇒ t = 30 ÷ 5

$\therefore$ t = 6 sec

Now,

⇒ Maximum velocity = ( 2 mt/s² )

= [ ( 2m × 6s ) / s² ]

= 12 m/s

Hence, the maximum velocity attained by the body is 12 m/s.