Question

An object starting from rest travels 20 m in the first 2 s and 160 m in

the next 4 s. What will be the velocity after 7 s from the start .​

Answer 1

❄️ Answer:-

Using second equation of motion:$~ \sf s = ut \: + \: \frac{1}{2} at {}^{2}$

Here :

s => distance covered = 20 m

a => acceleration = __ m/s²

u => initial velocity = 0 m/s

t => time = 2s

s => $~ \sf s = ut \: + \: \frac{1}{2} at {}^{2}$

$\\ \sf \twoheadrightarrow \: 20 = 0 + \frac{1}{2} \times a \times (2) {}^{2} \\ \\ \sf \twoheadrightarrow \:2a = 20 \\ \\ \sf \twoheadrightarrow \:a = 10 \: m/s²$

The final velocity, v after 2s

The first equation of motion: v = u + at

Here:

v => final velocity = ?

u => initial velocity = 0 m/s

a = acceleration = 10 m/s²

t = time = 2s

$\\ \large\sf \twoheadrightarrow \:v = u + at \\ \\ \sf \twoheadrightarrow \: v = 0 + 10 × 2 \\ \\ \sf \twoheadrightarrow \: v = 20 \: m/s²$

We know that the distance travelled in next 4 seconds is 160 m and the body is sta staring at 20 m/s.

$\sf \therefore \: u = 20 \: m/s² \\ \sf t = 4 \: sec \\ \sf s = 160 \: m$

Putting the above value in $~~ \sf \: s = ut + \dfrac{lat {}^{2} }{2}$

We get :- $\\ \: \: \sf160 = 20 \times 4 + \dfrac{la(4 {}^{2} )}{2}$

Solving for a we get a = 10 m/s²

Since the acceleration is same, we use

$\\ \sf \twoheadrightarrow \: v = u + at \\ \\ \sf \twoheadrightarrow \: v = 0 + 10 \times 7 \\ \\ \sf \twoheadrightarrow \: v = 70 \: m/s²$