Question

A car accelerates from 6 ms–1 to 16 ms–1 in 10

seconds. Calculate : (a) the acceleration and

(b) the distance covered by car in that time​

Answer 1

Answer:

Given that,

V=16m/s, u=6m/s

Time,t=10 seconds

A) From the first law of the motion,

V=u+at

16=6+10a

10a=10

a=1m/s²

B) from the second law of the motion,

S=ut+1/2at²

S=60+1/2×100

S=60+50

S=110m

Explanation:

Hope it helps you frnd..........

Answer 2

Answer

a ) 1 m/s²

b ) 110 m

Given

• A car accelerates from 6 m/s to 16 m/s in 10   seconds

To Find

a ) The acceleration

b ) The distance covered by car in that time

Solution

a )

We know that , " Acceleration is defined as rate of change of velocity "

$\implies \bf a=\dfrac{v-u}{t}\\\\\implies \bf a=\dfrac{16-6}{10}\\\\\implies \bf a=\dfrac{10}{10}\\\\\implies \bf a=1\ m/s^2$

b )

Apply 3rd equation of motion .

⇒ v² - u² = 2as

⇒ 16² - 6² = 2(1)s

⇒ 256 - 36 = 2 s

⇒ 220 = 2 s

⇒ s = 110 m

So , the distance covered by car in 10 sec is 110 m