Question

a car accelerates from 6ms-1 to 16ms-1 in 10 sec.calculate(a) the acceration and (b)the distance covered by the car in that time.

Answer 1

ATQ, a car accelerates from 6m/s to 16m/s in 10 seconds.

• initial velocity = v = 6m/s

• final velocity = u = 16m/s

• total time taken = t = 10 seconds
• acceleration = a = ?

➡ distance = s = ?

(a) using first equation of motion

➡ v = u + at

➡ 16m/s = 6m/s + 10a

➡ 16 - 6 = 10a

➡ 10m/s = 10a

➡ a = 10/10

➡ a = 1m/s²

hence, the acceleration of the car is 1m/s²

(b) now we can use two equation of motion 2nd and 3rd. answer will be same..

let's find the distance using both equations.

➡ s = ut + 1/2 at²

= (6 × 10) + 1/2 × 1 × 10²

= 60 + 1/2 × 100

= 60 + 50

= 110m

using 3rd equation

➡ 2as = v² - u² (here we don't need t = time)

➡ 2 × 1 × s = 16² - 6²

➡ 2s = 256 - 36

➡ s = 220/2

➡ s = 110m

hence, the distance covered by the car is 110m.

Answer 2

$\huge\underline\mathfrak{Answer\:-}$

Acceleration = 1 m/s²

Distance covered by the car = 110 m

________________________

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

Given :

Initial velocity (u) = 6m/s

Final velocity (v) = 16m/s

Time (t) = 10s

To find :

• Acceleration
• Distance covered by the car

Solution :

We know that,

V = u + at ( first equation of motion )

a = (v-u)/t

a = (16-6)/10

a = 10/10

a = 1m/s²

Hence, acceleration = 1m/s²

__________________________

Now,

We also know that,

S = ut + 1/2at² ( second equation of motion )

S = 6×10+1/2(1)(10)²

S = 60+1/2(100)

S = 60+50

S = 110 m

Hence, Distance = 110 m.

______________________________