Question

. A car starting from the rest moves with the uniform acceleration of 2 metre per second square for four minutes. Find (a)the speed acquired (b)the distance travelled. *​

Answer 1

Given:-

• Initial velocity,u = 0m/s

• Acceleration ,a = 2m/s²

• Time taken ,t = 4 min

To Find:-

• Final velocity,v

• Distance travelled ,s

Solution:-

Conversion of unit

1min = 60s

∴ 4min = 4×60 = 240s

Now using 1st equation of motion

→ v = u+at

Substitute the value we get

→ v = 0+ 2×240

→ v = 480m/s

∴ The final velocity of the car is 480m/s

Now Calculating the distance covered by the car So by using 2nd equation of motion

• s = ut +1/2at²

substitute the value we get

→ s = 0×240 + 1/2×2×(240)²

→ s = 0 + 240×240

→ s = 57600 m

→ s = 57600/1000 = 57.6km

∴ The Distance covered by the car is 57.6 km.

Answer 2

$\underline {\underline{\large{ \mathbb {ANSWER}}}}$

　　　　　　　　　　　　　　　　　　　　　　　

$\sf \color{gray} Given:-$

Initial velocity (u) = 0

Acceleration (a) = 2m/s²

Time taken (t) = 4min = 240s

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

$\sf \color {gray} To \: find:-$

(a) Final velocity (v)

(b) Distance travelled (s)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

$\sf \color{gray}Solution: -$

$\red{(a)} \: \: \boxed{ \rm {v = u+at}} </p><p>$

$⇢ \sf v = 0+2×240 </p><p>$

$</p><p> ⇢\sf v= 2×240 </p><p>$

$\therefore \: \sf \underline{\purple{v=480~ m/s}}$

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

$\red{(b)} \: \: \boxed {\rm{2as = {v}^{2} - {u}^{2} }}$

$⇢\sf2 \times 2 \times s = {480}^{2} - {0}^{2}$

$⇢\sf 4s = 230400 - 0$

$⇢ \sf 4s = 230400$

$⇢\sf s = \dfrac{230400}{4}$

$\therefore \: \sf \underline{\purple{s = 57600 \: m}}$