Question

A vehicle starts to move from test and attain an acceleration of 0.8 m/s in 10 sec. Calculate the final velocity and distance covered by the vehicle within that time.

Answer 1

Answer :-

• final velocity of vehicle = 8 m/s
• distance covered by vehicle = 40 m.

Given :-

• initial velocity of vehicle , u = 0
• acceleration attained by vehicle , a = 0.8 m/s
• time taken to attain acceleration , t = 10 s

To find :-

• final velocity of vehicle , v = ?
• distance covered by vehicle within the time , s = ?

Formula required :-

• First equation of motion

      v = u + a t

• second equation of motion

     s = u t + 1/2 a t²

[ where v is final velocity , u is initial velocity , a is acceleration , s is distance covered and t is time taken ]

Calculation :-

Using first equation of motion

→ v = u + a t

→ v = 0 + (0.8) (10)

→ v = 8 m/s

therefore,

final velocity of vehicle will be 8 m/s.

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = (0) (10) + 1/2 (0.8) (10)²

→ s = (0.4) (100)

→ s = 40 m

therefore,

Distance covered by vehicle within 10 sec is 40 m .

Answer 2

$\sf\large{\underline{Question:-}}$

A vehicle starts to move from test and attain an acceleration of 0.8 m/s in 10 sec. Calculate the final velocity and distance covered by the vehicle within that time.

$\sf\large{\underline{Given:-}}$

• Initial velocity u = 0
• acceleration a = 0.8 m/s
• time t = 10 sec

$\sf\large{\underline{Find:-}}$

• finial velocity v = ?
• Distance s = ?

$\sf\large{\underline{Solution:-}}$

• using 1st equation of motion

→ v = u + at

→ v = 0 + 0.8*10

→ v = 8 m/s

Therefore ,

• Final velocity of vehicle = 8 m/sec

Now,.

• calculating distance travelled by vehicle

• using 2nd equation of motion

→ s = ut + 1/2at²

→ s = 0*10 + 1/2* 0.8 * 10²

→ s = 0.4 * 100

→ s = 40m/sec

Therefore ,

• Distance travelled by vehicle = 40m/sec