Question

A car moving with uniform acceleration at a veloctiy 32 m /s from 16 m/s after 4 seconds calculate its acceleration and distance travelled by this time​

Answer 1

Given :

• Intial velocity = 16 m/s.

• Final Velocity = 16 m/s.

• Time = 4 s.

To find :

• Acceleration of the car.

• Distance covered by the car.

Solution :

To find the Acceleration of rhe car :

We know the first Equation of Motion i.e,

⠀⠀⠀⠀⠀⠀⠀⠀⠀v = u + at

Where :

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Now using the first Equation of Motion and substituting the values in it, we get :

➠ v = u + at

➠ 32 = 16 + a × 4

➠ 32 - 16 = 4a

➠ 16 = 4a

➠ 16/4 = a

➠ 4 = a

∴ a = 4 m/s².

Hence the Acceleration of the car is 4 m/s².

To find the distance covered :

Method (i) :

We know the second Equation of Motion i.e,

⠀⠀⠀⠀⠀⠀⠀S = ut + ½at²

Where :

• S = Distance
• u = Intial velocity
• a = Acceleration
• t = Time Taken

Using the second Equation of Motion and substituting the values in it , we get :

➠ S = ut + ½at²

➠ S = 16 × 4 + ½ × 4 × 4²

➠ S = 64 + ½ × 4 × 16

➠ S = 64 + ½ × 64

➠ S = 64 + 32

➠ S = 96

∴ S = 96 m.

Hence the distance covered by the car is 96 m.

Method (ii) :

We know the third Equation of Motion i.e,

⠀⠀⠀⠀⠀⠀⠀⠀v² = u² + 2aS

Where :

• v = Final Velocity
• u = Intial velocity
• a = Acceleration
• S = Distance

Using the third Equation of Motion and substituting the values in it, we get :

➠ v² =u² + 2aS

➠ 32² = 16² + 2 × 4 × S

➠ 1024 = 256 + 2 × 4 × S

➠ 1024 - 256 = 2 × 4 × S

➠ 768 = 2 × 4 × S

➠ 768 = 8 × S

➠ 768/8 = S

➠ 96 = S

∴ S = 96 m.

Hence the distance covered by the car is 96 m.

Answer 2

Given

• Initial velocity (u) = 16 m/s
• Final velocity (v) = 32 m/s
• Time (t) = 4 sec

To find

• Acceleration and distance travelled.

Solution

★ By using first equation of motion,

$\underline{\boxed{v = u + at}}$

→ 32 = 16 + a × 4

→ 32 = 16 + 4a

→ 32 - 16 = 4a

→ 4a = 16

→ a = 4

•°• a = 4 m/s²

★ Now, we can calculate distance by using second equation of motion

$\underline{\boxed{s = ut + \dfrac{1}{2} × at^2}}$

→ s = 16 × 4 + ½ (4)(4)²

→ s = 64 + ½ × 64

→ s = 64 + 32

→ s = 96 m

Hence, the distance travelled by car is 96 m.