Question

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s covering a distance of 79.8m. Calculate the time taken to change the velocity.

Answer 1

$\huge\bold\red{HELLO!}$

Hello,

Find your answer below.

u=18.5 m/s

v=46.1 m/s

t=2.47 secs                                                                                                                 

a=(v-u)/t

= (46.1-18.5)/ 2.47

= 27.6/2.47= 11.17m/s^2

So, the accelaration is 11.17 m/s^2

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Answer 2

Answer:

$\huge \boxed{\bold{2.47\ seconds}}$

Question

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s covering a distance of 79.8m. Calculate the time taken to change the velocity.

Given

$V_{i} = 18.5\ m/s\\V_{f} = 46.1\ m/s\\x = 79.8\ m\\t = ?$

Formula

Let us use the formula $x = \frac{1}{2} (V_{f}+ V_{i}) t$ and derive $t$.

$x = \frac{1}{2} (V_{f}+ V_{i}) t\\\frac{1}{\frac{1}{2}(V_{f}+V_{i})} (x) = (\cancel{\frac{1}{2} (V_{f}+ V_{i})} t) \frac{1}{\cancel{\frac{1}{2}(V_{f}+V_{i})}}\\\bold{t = \frac{x}{\frac{1}{2}(V_{f}+V_{i})}}$

Solution

[Sol]

Substitute the given values to the derived formula.

$t = \frac{x}{\frac{1}{2}(V_{f}+V_{i})}\\= \frac{79.8\ m}{\frac{1}{2}(46.1\ m/s\ + 18.5\ m/s)}\\= \frac{79.8\ m}{\frac{1}{2}(64.6\ m/s)}\\= \frac{79.8\ \cancel{m}}{32.3\ \cancel{m}/s}\\= \bold{2.470\ s}$

From the solution above, the time the car had taken to change the velocity is 2.47 seconds.

~Hope it helps.~

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