Question

*PLZ HELP IT'S DUE IN 2HRS*
So let's say a cart (m= 16 kg) is currently traveling with a velocity of 3 m/s.
The cart then gains speed, achieving a new velocity of 14 m/s after 22 seconds.

What would be the calculation of the cart's acceleration during the 22 second interval and how?
A.) + 44 m/s^2
B.) 8 m/s^2
C.) +0.5 m/s^2
D.) -0.5 m/s^2
E.) + 11 m/s^2

Answer 1

Answer:-

$\bigcirc$$\sf{44m/s {}^{2}}$

$\bigcirc$$\sf{8m/s {}^{2}}$

$\odot$$\sf{0.5m/s {}^{2}}$

$\bigcirc$$\sf{-0.5m/s {}^{2}}$

$\bigcirc$$\sf{11m/s{}^{2}}$

Calculations:-

$\sf Given\begin{cases}\sf {Mass_(m)=16kg} \\ Initial\: velocity_(u)=3m/s \\ Final\;velocity_(v)=14m/s \\ time\:taken_(t)=22s \end {cases}$

$\sf To\: find\begin {cases}Acceleration_(a)\end {cases}$

Solution:-

• According to first equation of kinematics

${\boxed{\sf {v=u+at}}}$

• Substitute the values

${:}\longmapsto$$\sf {14=3+a×22}$

${:}\longmapsto$$\sf {14=3+22a}$

${:}\longmapsto$$\sf {22a=14-3 }$

${:}\longmapsto$$\sf {22a=11 }$

${:}\longmapsto$$\sf {a={\dfrac {1}{2}}}$

${:}\longmapsto$${\underline{\boxed{\bf {a=0.5m/s {}^{2}}}}}$

$\therefore$$\sf {The\:acceleration\:during\:22seconds\:of\:interval\:is} \\ \sf {0.5m/s {}^{2}.}$

Answer 2

Given :

• Initial velocity (u) = 3 m/s
• Final velocity (v) = 14 m/s
• Time (t) = 22 seconds
• Mass of a cart (m) = 16 kg

To find :

• Acceleration (a)

According to the question,

By using Newtons first equation of motion,

→ v = u + at

Where,

• v stands for Final velocity
• u stands for Initial velocity
• a stands for Acceleration
• t stands for Time

→ Substituting the values,

→ 14 = 3 + a × 22

→ 14 - 3 = 22a

→ 11 = 22a

→ 11 ÷ 22 = a

→ 0.5 = a

So,the acceleration of the cart is 0.5 m/s².

.°. Option (D) 0.5 m/s² is correct.

__________________________

More formulas :

★ Newton's first equation of motion :

• v = u + at

★ Newton's second equation of motion :

• s = ut + ½ at²

★ Newton's third equation of motion :

• v² = u² + 2as