Question

A car is travelling with 54 km/h,when breaks is applied it stops in 5 second, what will be acceleration and distance it will cover before it stop.

Answer 1

Answer:

• Acceleration = 3m/s
• Distance covered = 37.5

Explanation:

We know,

→ Acceleration = (v - u)/t

Where, v denotes the final velocity, u denotes the initial velocity, and t is the time taken.

We are given with,

• v = 0km/h or, 0m/s (since, it comes to rest)
• u = 54km/h or, 15m/s (By conversion)
• t = 5s

From the given parameters,

→ (v - u)/t

→ (15 - 0)/5

→ 15/5

→ 3m/s

∴ Acceleration of the car is 3m/s

Now,

Calculating the distance covered.

→ s = u + ½.a.t²

Where,

• u(initial velocity) = 0m/s
• a(acceleration) = 3m/s
• t(time) = 5s

Then,

→ s = 0 + ½(3)(5)²

→ s = ½.3.25

→ s = 37.5m

∴ Distance covered by the car is 37.5m

Answer 2

Answer :-

a = -3 m/s²

d = 37.5 m

Explanation :-

according to first equation of motion

a = v - u /t

where,

a = acceleration

v = final velocity

u = initial velocity

and, t = time

____________________________

Now, given that

u => 54 km/h = 15 m/s

v => 0 km/h = 0 m/s

t => 5 sec

therefore, a =

$\frac{0 \: - 15 \: }{5} \\ = \frac{ - 15}{5}$

a => -3 m/s²

___________________________

according to second equation of motion,

s = ut² + 1/2 at²

where,

s = distance

u = initial velocity

a = Acceleration

t = time

____________________________

u = 0 m/s

a = -3 m/s²

t = 5sec

then, s =

$= > 0 \times {5}^{2} + \frac{1}{2} ( 3) \times {5}^{2} \\ = > 0 + \frac{75}{2}$

therefore s = 37.5 m

Hence, acceleration = -3m/s²

and distance = 37.5 m