Question

A car starting from rest travels for 5s covering a displacement of 50m. It’s constant acceleration is

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Displacement ,s = 50m
• Time taken ,t = 5 s

To Find:-

• Acceleration ,a

Solution:-

According to the given statement !!

It is given that the car start from rest and travel 50 m in 5 second. We have to calculate the acceleration of the car . Using 2nd equation of motion we get

• s = ut + 1/2at²

where,

• s is the displacement
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

$:\implies$ 50 = 0×5 + 1/2×a × 5²

$:\implies$ 50 = 0 + 1/2× a × 25

$:\implies$ 50×2 = 25×a

$:\implies$ 100 = 25×a

$:\implies$ 100/25 = a

$:\implies$ a = 100/25

$:\implies$ a = 4 m/s²

• Hence, the acceleration of the car is 4 m/s².

Answer 2

Answer:

Given :-

• A car starting from rest travels for 5 seconds a displacement of 50 m.

To Find :-

• What is the constant acceleration.

Formula Used :-

$\clubsuit$ Second equation of motion :

$\longmapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

where,

• s = Distance travelled or Displacement
• u = Initial Velocity
• t = Time
• a = Acceleration

Solution :-

Given :

• Initial Velocity (u) = 0 m/s
• Distance travelled or Displacement (s) = 50 m
• Time (t) = 5 seconds

According to the question by using the formula we get,

$\dashrightarrow \sf 50 =\: (0)(5) + \dfrac{1}{2} \times a(5)^2$

$\dashrightarrow \sf 50 =\: 0 + \dfrac{1}{2} \times a(25)$

$\dashrightarrow \sf 50 =\: 0 + \dfrac{1}{2} \times 25a$

$\dashrightarrow \sf 50 - 0 =\: \dfrac{1}{2} \times 25a$

$\dashrightarrow \sf 50 =\: \dfrac{1}{2} \times 25a$

$\dashrightarrow \sf 50 \times 2 =\: 25a$

$\dashrightarrow \sf 100 =\: 25a$

$\dashrightarrow \sf a =\: \dfrac{\cancel{100}}{\cancel{25}}$

$\dashrightarrow \sf a =\: \dfrac{4}{1}$

$\dashrightarrow \sf\bold{\red{a =\: 4\: m/s^2}}$

$\therefore$ The constant acceleration is 4 m/s².

$\rule{150}{2}$

EXTRA INFORMATION:

$\clubsuit$ First equation of motion :

$\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Second equation of motion :

$\longmapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

$\clubsuit$ Third equation of motion :

$\longmapsto \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}$

where,

• s = Distance Travelled
• u = Initial Velocity
• v = Final Velocity
• a = Acceleration
• t = Time