Question

A racing car has a uniform acceleration of 4m/52s. What distance will it cover in 10s after the car starts

Answer 1

Answer:

Uniform acceleration a = 4 m/s²

Initial velocity u= 0 m/s

Time t = 10 sec.

Velocity v after 10 sec= u + a t = 0 m/s + 4 m/s² × 10 s =40 m/s.

We shall calculate the distance covered by all possible methods

Using Average Velocity:

Initial velocity u= 0 m/s

Final velocity v= 40 m/s

Average velocity= ½( u +v) = ½ ( 0+40)= 20 m/s

Time= 10 s

Distance covered in 10 s= av. vel. × time= 20m/s × 10s = 200 m

Using the relation s = u t+ ½ a t²

u = 0 m/s, a = 4 m/s², t= 10 sec.

s = 0×10 + ½ 4× 10² = 200 m

Using the relation v² - u² = 2 a s,

u = 0 m/s, v= 40 m/s, a= 4 m/s²

s = 1/2 a ( v² - u²)= 1/2×4 ( 40² - 0²) = 1/8(1600–0)= 1600/8= 200 m.

So all relations give the same results as it should

Answer 2

$\boxed {Your \: Answer}$

$Acceleration$ = $4$$\frac{m}{s}^{2}$

$Initial$ $Velocity$ = $0$ $\frac{m}{s}$

$Time$ = $10$ $Seconds$

$Distance$ $Covered$ $(s)$ = ?

$\underline\text{We\:\:Know\:\:distance(s):-}$ = $ut$ + $\frac{1}{2}$${at}^{2}$

➝ $s$ $=$ $0$ $×$ $10$ $+$ $\frac{1}{2}$ $×$ $4$ $×$ ${10}^{2}$$m$

➝$s$ $=$ $2$ $×$ $100$$m$

➝$s$ $=$ $200$$m$

$\underline\text{Thus\:\:racing\:\:car\:\:will\:\:cover\:\:a\:\:distance\:\:of\:\:200m\:\:after\:\:start\:\:in\:\:10s\:\:with\:\:given\:\:acceleration:-}$