Question

Beginning from rest, Batman accelerates his Batmobile to reach a velocity of 60 meters per second in 10 seconds. Then he applies the brakes and the velocity of the Batmobile decreases to 10 meters per second in the next 1 second. Calculate the acceleration of the Batmobile in both cases.​

Answer 1

Answer:

• Acceleration in 1st case = 6m/s²
• Acceleration in 2nd case = -50m/s² (negative sign show retardation).

Explanation:

Solution:-

As it is given in 1st condition that the

• initial velocity ,u = 0m/s
• Final Velocity ,v = 60m/s
• Time taken ,t = 10 s

we have to calculate the acceleration of the batman

Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

substitute the value we get

→ a = 60-0/10

→ a = 60/10

→ a = 6m/s²

• Hence, the acceleration in 1st case is 6m/s².

Now, calculating acceleration in 2nd case .

In second case the is given

• Initial velocity ,u = 60m/s
• Final Velocity ,v = 10m/s
• Time taken ,t = 1

Acceleration in 2nd case ?

• a' = v-u/t

→ a' = 10-60/1

→ a' = -50/1

→ a' = -50m/s²

Here, negative sign show retardation

• Hence, the acceleration of the batmobile in 2nd case is 50m/s².

Answer 2

Answer:

1]

Given :-

• Initial velocity = 0 m/s
• Final Velocity = 60 m/s
• Time = 10 sec

To Find :-

Acceleration

Solution :-

We know that

$\large \sf \: a \: = \dfrac{v - u}{t}$

a = Acceleration

v = Final Velocity

u = Initial Velocity

t = Time

$\sf \: a = \dfrac{60 - 0}{10}$

$\sf \: a \: = \dfrac{60}{10}$

$\sf \: a \: = 6 \: {m/s}^{2}$

Now,

2]

Given :-

• Initial velocity = 60 m/s
• Final Velocity = 10 m/s
• Time = 1 sec

To Find :-

Acceleration

Solution :-

$\large \sf \: a \: = \dfrac{v - u}{t}$

a = Acceleration

v = Final Velocity

u = Initial Velocity

t = Time

$\sf \: a \: = \dfrac{10 - 60}{1}$

$\sf \: a \: = \dfrac{ - 50}{1}$

$\sf \: a \: = - 50 \: m/s^{2}$