Question

Topic - Kinematics
A bus starts moving with uniform acceleration from its position of rest. It moves 48 m in 4 s. On applying the brakes, it stops after covering 24 m. Find the deceleration of the bus.​

Answer 1

QUESTION-:

A bus starts moving with uniform acceleration from its position of rest. It moves 48 m in 4 s. On applying the brakes, it stops after covering 24 m. Find the deceleration of the bus.​

EXPLANATION-:

• Initial velocity(u)= 0 m/s
• Final velocity(v)= ??
• Distance(s)=24 m
• Acceleration(a)= ?

First let's calculate the acceleration  in first 4 s of the motion

Using 2nd equation of kinematics-:

$\bf \red\bigstar \: \: \orange{ \underbrace{ \underline{\tt \blue{ s== ut + \frac{1}{2}at^2}}}}$

Putting values-:

→48=0×4+1/2×a×(4)²

→48=1/2×a×16

→48=8a

→a=48/8

$\dashrightarrow \underline{\underline{\boxed{\bf\; a=6\;m/s^2}}}$

Since we know that acceleration,we can calculate the final velocity of bus.

Using 1st equation of kinematics-:

$\bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \tt\blue{ v = u + at}}}}$

Putting values-:

→v=0+6(4)

→v=6×4

$\dashrightarrow \underline{\underline{\boxed{\bf\; v=24\;m/s}}}$

[NOTE-: NOW AFTER 4 s THE FINAL VELOCITY WILL BECOME THE INITIAL VELOCITY AND AT END THE FINAL VELOCITY WILL BE ZERO BECAUSE THE BUSS STOPS ]

Now let's calculate the acceleration from starting-:

Using 3rd equation of motion-:

$\bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \tt\blue{ v^2=u^2+2as}}}}$

Putting values:-

→(0)²=(24)²+2(a×24)

→0=576+48a

→-576=48a

→a=-576/48

$\dashrightarrow \underline{\underline{\boxed{\bf\; a=-12\;m/s^2}}}$

We know that-:

$\longrightarrow \boxed{\underline{\tt\; Deceleration-(acceleration)}}$

Putting values-:

→deceleration =-(-12)

$\dashrightarrow \underline{\underline{\boxed{\bf\;deceleration=12\;m/s^2}}}$

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