Question

A ball starting from the rest hits the wall with the velocity 40 m/s. Find the distance covered by the ball in 4 seconds.

(1 Point)

40 m

160m

80m

120m


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Answer 1

Answer:

80m

Explanation:

initial velocity= 0

final velocity= 40m/s

acceleration= v-u/t

= 40-0/4

= 10m/s^-2

v^2-u^2/2a=x

40^2-0/2×10=x

1600/20=x

x=80m

Answer 2

Answer :

➥ The Distance covered by the ball = 160 m

Given :

➤ Intial velocity of a ball (u) = 0 m/s

➤ Final velocity of a ball (v) = 40 m/s

➤ Time taken by a ball (t) = 4 sec

To Find :

➤ Distance covered by the ball (s) = ?

Solution :

To find the distance covered by the ball, first we need to find the acceleration of the ball, then after, we will find the distance covered by the ball.

We can find Acceleration by using the first equation of motion which says v = u + at.

☁️ So let's calculate a !

$\tt{\hookrightarrow v = u + at}$

$\tt{\hookrightarrow 40 = 0 + a \times 4}$

$\tt{\hookrightarrow 40 = 0 + 4a}$

$\tt{\hookrightarrow 40 = 4a}$

$\tt{\hookrightarrow \cancel{\dfrac{40}{4}} = a }$

$\tt{\hookrightarrow 10 = a}$

$\bf{\hookrightarrow \underline{ \: \: \underline{ \orange{ \: \: a = 10 \: m/s^2 \: \: }} \: \: }}$

Now, we have Initial velocity, time taken, and it's Acceleration of the ball,

• Intial velocity of the ball (u) = 0 m/s
• Time taken by the ball (t) = 4 sec
• Acceleration of the ball (a) = 10 m/s²

☃️ So let's find the distance covered by the ball in 4 seconds by using the formula:

From second equation of motion

$\tt{:\implies s = ut + \dfrac{1}{2}at^2}$

$\tt{:\implies s = 0 \times \dfrac{1}{2} \times 10 \times 4^2}$

$\tt{:\implies s = 0 \times \dfrac{1}{\cancel{2}} \times 10 \times \cancel{4} \times 4}$

$\tt{:\implies s = 0 + 1 \times 10 \times 2 \times 4}$

$\tt{:\implies s = 0 + 10 \times 2 \times 4}$

$\tt{:\implies s = 0 + 20 \times 4}$

$\tt{:\implies s = 0 + 160}$

$\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: s = 160 \: m \: \: }} \: \: }}$

║Hence, the distance covered by the ball is 160 m.║