Question

A racing car has a uniform acceleration of 4 ms-1 what distance will it cover in 10 seconds after start

Answer 1

Answer:

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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

➭ Question :-

$⟶$A racing car has a uniform acceleration of 4 m/s² what distance will it cover in 10 seconds after start

➭ Given:-

$⟶$Initial velocity is:- u = 0

$⟶$Acceleration is:- a = 4 m/s²

$⟶$Time is:- t = 10 s

➭ To find :-

$⟶$the distance covered in 10second after start

➭ Formula required :-

$⟶$$\boxed{\sf S = ut + \frac{1}{2} {at}^{2} }$

➭ Solution :-

using 2nd equation of motion

$⟹ s = ut + \frac{1}{2} {at}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ⟹ s = 0 \times 10 + \frac{1}{2} \times 4 \times {10}^{2} \\ ⟹ 0 + 200 = 200m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence, 200m will be covered in 10sec after start