Question

A bus accelerates uniformly from 100mper second to 220metre per second in 1 minutes calculate a.the acceleration of the car b distance covered by the car in the interval

Answer 1

Appropriate Question:

A bus accelerates uniformly from 100 m/s to 220 m/s in 1 minute. Calculate

a) The acceleration of the car.

b) Distance covered by the car in the interval.

Answer:

Acceleration = 2 m/s²

Distance = 9600 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 100 m/s
• Final velocity (v) = 220 m/s
• Time (t) = 1 minute = 60 s

We have been asked to calculate acceleration and the distance covered by the car in the interval.

$\large {\underline { \sf {Calculating \; Acceleration :}}}$

❝ Acceleration is nothing the but the rate of change in velocity. ❞ It is a vector quantity and its SI unit is m/s². As I said that acceleration rate of change in velocity. So, we can write in the form of equation,

$\dashrightarrow \quad \rm { Acceleration= \dfrac{Change \; in \; velocity}{Time} }$

• Change is velocity can also be defined as the difference of the final velocity and the initial velocity.

$\dashrightarrow \quad \rm { Acceleration= \dfrac{Final \; Velocity - Initial \; Velocity}{Time} }$

• Acceleration is denoted by a
• Final velocity is denoted by v
• Initial velocity is denoted by u
• Time is denoted by t

So,

$\bigstar \quad \underline{\boxed{ \bf {a = \dfrac{v-u}{t} }}}$

Substitute the value in the formula.

$\dashrightarrow \quad \rm {a = \Bigg ( \dfrac{220 - 100}{60} \Bigg ) \; m \: s^{-2} }$

$\dashrightarrow \quad \rm {a = \Bigg ( \dfrac{120}{60} \Bigg ) \; m \: s^{-2} }$

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

∴ The acceleration of the car is 2 m/s².

$\rule{200}2$

$\large {\underline { \sf {Calculating \; Distance \; Travelled :}}}$

We can calculate the distance travelled easily by using the third equation of motion.

Third equation of motion:

$\bigstar \quad \underline{\boxed{ \bf {v^2 - u^2 = 2as }}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\dashrightarrow \quad \rm {(220)^2 - (100)^2 = 2 \times 2 \times s }$

$\dashrightarrow \quad \rm {48400 - 10000 = 4 \times s }$

$\dashrightarrow \quad \rm {38400 = 4s }$

$\dashrightarrow \quad \rm {\cancel{\dfrac{38400}{4}} = s }$

$\dashrightarrow \quad \underline{\boxed{ \bf {s = 9600 \; m }}}$

∴ The distance travelled by the bus is 9600 m.

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

A bus accelerates uniformly from 100 m/s to 220 m/s in 1 minute . Calculate

• a. The acceleration of the bus .

• b. Distance covered by the bus in the interval .

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

• a. Acceleration of the bus = 2 m/$\tt {s}^{2}$

• b. Distance Covered by bus = 9600 m

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

$\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

• Initial velocity ( u ) = 100 m/s

• Final velocity ( v ) = 220 m/s

• Time taken ( t ) = 1 minute = 60 seconds

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

• a. The acceleration of the bus .

• b. Distance covered by the bus in the interval .

$\green{ \large \underline{ \mathbb{\underline{ EQUATIONS \: OF \: MOTION \: USED : }}}}$

$\purple{ \boxed{\begin{array}{l} \sf v = u + at \\ \\ \sf {v}^{2} - {u}^{2} = 2as \: \: \end{array}}}$

$\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}$

$\large{\purple{ \underline{\underline{\textsf{a. Acceleration of the bus}}}}}$

$\displaystyle \sf \longrightarrow v = u + at \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 220 = 100 + a(60) \\\ \\ \displaystyle \sf \longrightarrow 220 = 100 + 60a \: \: \: \\ \\ \displaystyle \sf \longrightarrow 100 + 60a = 220 \: \: \: \\ \\ \displaystyle \sf \longrightarrow 60a = 220 - 100 \: \: \\ \\ \displaystyle \sf \longrightarrow 60a = 120 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow a = \frac{120}{60} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow a = 2 \: m {s}^{ - 2} \: \: \: \: \: \: \: \: \: \: \: \:$

$\large{\purple{ \underline{\underline{\textsf{b. Distance Covered by bus}}}}}$

$\displaystyle \sf \longrightarrow {v}^{2} - {u}^{2} = 2as \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow {(220)}^{2} - {(100)}^{2} = 2(2)s \: \\ \\ \displaystyle \sf \longrightarrow 48400 - 10000 = 4s \: \: \: \: \: \: \: \: \\ \\\displaystyle \sf \longrightarrow 38400 = 4s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 4s = 38400 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = \frac{38400}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = 9600 \: m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{\boxed{ \begin{array}{l} \textsf{a. Acceleration of the bus = 2 m { \sf s}^{ - 2} } \\ \\ \textsf{b. Distance Covered by bus = 9600 m} \end{array}}}$

$\green{ \large \underline{ \mathbb{\underline{KNOW\:MORE: }}}}$

• Acceleration

Acceleration is the rate at which velocity changes with time .

• Initial Velocity

Initial velocity is the velocity of the object before the effect of acceleration .

• Final Velocity

Final velocity is the velocity of the object after the effect of acceleration .

• Distance

Distance is the length of actual path covered by a moving object in a given time interval .

$\Large{\purple{\boxed{\begin{array}{l} \textsf{Equations of motion : } \\ \\ \textsf{v = u + at} \\ \\ \displaystyle\textsf{s = ut + \sf\frac{1}{2}a {t}^{2} } \\ \\ \sf {v}^{2} - {u}^{2} = 2as \end{array}}}}$