Question

A racing car has a uniform acceleration of 4 m s - ′2. What distance will it cover in 10 s after start?

Answer 1

Answer:-

Known Terms:-

u = Initial Velocity

a = Acceleration

t = Time

s = Distance

Given:-

Initial Velocity of  car, u = 0 ms-1

Acceleration of car, a = 4 m s-2

Time taken by the car, t = 10 s

To Find:-

Distance (s)

Solution:-

Here we have to find the distance

so,

Here, we use 3rd equation of motion that is  s = ut + $\frac{1}{2}$at²

⇒ s = ut + $\frac{1}{2}$at²

⇒ s = 0 + ($\frac{1}{2}$) × 4 × 10²

⇒ s = 0 + ($\frac{1}{2}$) × 4 × 10 × 10

⇒ s = ($\frac{1}{2}$) × 400 m

⇒ s = 200 m

The distance cover in 10 s after start is 200 m.

Answer 2

ANSWER: ❷⓪⓪m

 ⒺⓍⓅⓁⒶNⒶⓉⒾⓄⓃ 

$\huge\text{\underline{\underline{GIVEN}}}$

Initial Velocity of  the car $\red{\boxed{\mathfrak{ (u) = 0 ms^{-1} }}}$

Acceleration of the car  $\red{\boxed{\mathfrak{ (a) = 4 ms^{-2} }}}$

TIME TAKEN $\red{\boxed{\mathfrak{ (t) = 10s}}}$

WE HAVE TO FIND $\red{\boxed{\mathtt{DISTANCE (s)}}}$

$\huge\text{\underline{\underline{SOLUTION}}}$

To find the distance let's use the 2nd equation of motion

WHICH IS $\red{\boxed{\mathfrak{ s = ut + \frac{1}{2}at^2}}}$

$\red{\mathfrak{ \implies s = ut + \frac{1}{2}at^2}}}$

NOW PUTTING THE VALUES

$\red{\mathfrak{ \implies s = 0 + (\frac{1}{2}) \times 4 \times 10^2}}}$

$\red{\mathfrak{ \implies s = 0 + (\frac{1}{2}) \times 4 \times 100}}}$

$\red{\mathfrak{ \implies s = (\frac{1}{2}) \times 400 m}}}$

$\huge{\red{\boxed{\mathfrak{ \implies s = 200 m}}}}$

∴The CAR will cover 200 m distance after 10 s.