Question

An object starts from rest and accelerates at the rate of 2m/s2 in 4 s.What distance will it cover? ​

Answer 1

Answer:

4m/s

Explanation:

Let v,t  

1

​

,t  

2

​

 be the maximum velocity attained, time for which car accelerated and the time for which car retarded respectively

⇒t  

1

​

=(v−u)/a=(v−0)/2=v/2

⇒t  

2

​

=(v−u)/a=(0−v)/(−4)=v/4

Now it is given that,  

⇒t  

1

​

+t  

2

​

=3

⇒v/2+v/4=3

⇒v=4m/s

Answer 2

Answer:

Given :-

• An object starts from rest and accelerates at the rate of 2 m/s² in 4 seconds.

To Find :-

• What is the distance it will cover.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

Solution :-

Given :

• Initial Velocity = 0 m/s [Object starts from rest]
• Acceleration = 2 m/s²
• Time Taken = 4 seconds

According to the question by using the formula we get,

$\longrightarrow \sf s =\: (0)(4) + \dfrac{1}{2} \times (2)(4)^2$

$\longrightarrow \sf s =\: 0 \times 4 + \dfrac{1}{2} \times 2 \times 4 \times 4$

$\longrightarrow \sf s =\: 0 + \dfrac{1}{\cancel{2}} \times {\cancel{32}}$

$\longrightarrow \sf s =\: 0 + 16$

$\longrightarrow \sf\bold{\red{s =\: 16\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: it\: will\: cover\: is\: 16\: m\: .}}}}$