Question

• A vehicle change its speed from 68m/s to 16 m/s im 8 sec. Calculate acceleration and
distance covered in that time.​

Answer 1

Answer:

Given,

Initial Velocity (u) = 68 m/s

Final Velocity (v) = 16 m/s

Time (t) = 8 s

i. Acceleration = (v-u)/t

= (16-68)/8 m/s²

= -6.5 m/s²

Retardation = 6.5 m/s²

ii. Distance Covered (S) = ut-0.5at²

= (68*8) - [ 0.5 * -6.5 * 8 * 8 ] m

= 544 - 208 m

= 336 m or, 0.336 km

Answer: The Acceleration is -6.5 m/s² or Retardation = 6.5 m/s² and the Distance Travelled is 336 m or, 0.336 km.

I hope this helps.

Answer 2

Answer :-

Acceleration = - 6.5 m/$s^2$

Distance covered in that time = 336 m

To find :-

• Acceleration

• Distance covered in given time

Solution :-

Here,  

Initial Velocity (u) = 68 m/s

Final Velocity (v) = 16 m/s

Time (t) = 8 seconds

We will find acceleration from the first equation of motion which is

$\longrightarrow$ v = u + at

where,

Final Velocity = v

Initial Velocity = u

Time = t

Acceleration = a

Substituting the values in the equation

=> 16 m/s = 68 m/s + (a * 8 s)

=> 16 m/s - 68 m/s =  a * 8 s

=> - 52 m/s =  a * 8 s

=> $\dfrac{- 52\:m/s}{8\:s}$ = a

=> - 6.5 m/$s^2$ = a

Acceleration is negative which indicates that it is retardation.

We will find the distance from the second equation of motion which is

$\longrightarrow$ S = ut + $\dfrac{1}{2} at^{2}$

where,

Distance = S

Substituting the values in the equation

=> S = (68 m/s * 8 s) +( $\dfrac{1}{2} \times -6.5 m/s^2 \times (8 s)^{2}$)

=> S = 544 m - 208 m

=> S = 336 m

Distance covered in that time = 336 m