Question

a truck starting from rest moves down a hill with a constant acceleration.if it covers a distance of 300m in 10s find its acceleration

Answer 1

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

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

u = 0 m/a.

S = 300 m.

t = 10 seconds.

a = ?.

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

As "u" , "S" , "t" are given and need to find acceleration.

So, we have to use Second kinematic equation.

$\large{\bold{S = ut + \frac{1}{2} at^2}}$

Substituting the values.

$\large{ \to300 = 0 + \frac{1}{2} a \times {10}^2}$

$\large{ \to 300 \times 2 = 100 \times a}$

$\large{ \to 600 = a \times 100}$

$\large{ \to a = \frac{600}{100}}$

$\large{ \to a = 6 m/s^2}$

$\huge{\boxed{\boxed{a = 6 m/s^2}}}$

So,the acceleration of the truck is 6 m/s²

Answer 2

$\huge{\underline{\underline{\sf{Answer \colon}}}}$

From the Question,

As the object is starting from rest

• Initial Velocity,u = 0m/s

• Distance Travelled,s = 300m

• Time taken,t = 10s

Using the relation,

$\sf{s = ut + \frac{1}{2}at {}^{2}} \\ \\ \implies \: \boxed{ \sf{s = \frac{1}{2}at {}^{2} }}$

Substituting the appropriate values,we get:

$\sf{300 = \frac{1}{2}a(10) {}^{2} } \\ \\ \implies \: \sf{\frac{a}{2} = 3} \\ \\ \implies \: \huge{ \sf{a = 6ms {}^{ - 2} }}$

The truck moves with a constant acceleration of 6m/s²