Question

A car start from rest and covers a distance of 40m in 10s.calculate the magnitude of it acceleration

Answer 1

Answer:

0.8 m / sec² .

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Distance ( s ) = 40 m

Time ( t ) = 10 sec

We have to find acceleration :

Using second equation of motion :

s = u t + 1 / 2 a t²

Putting given values here we get

40 = 0 + 1 / 2 × a × 10 × 10

40 = 5 ×  a × 10

5 a = 4

a = 0.8 m / sec² .

Thus the acceleration of car is 0.8 m / sec² .

Answer 2

$\huge{\underline{\underline{.........Answer..........}}}$

$\huge{\underline{Given:-}}$

u = 0 m/s.

S = 40 m.

t = 10 seconds.

$\huge{\underline{Explanation:-}}$

As "u" ,"s" and "t" are given then should apply second kinematic equation .

From second kinematic equation.

$\huge{\boxed{\boxed{S = ut + \frac{1}{2} a {t}^{2}}}}$

substituting the values

${40 = 0 + \frac{1}{2} \times a \times {10}^{2}}$

${a = \frac{80}{100}}$

${a = 0.8 m/s^2}$

$\huge{\boxed{\boxed{a = 0.8 m/s^2}}}$

so,the acceleration of car is 0.8 m/s².