A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 ms^{-1}$ to $20ms^{-1}$ while passing through a distance 135 m in t second. The value of t is

Question

A car starts from rest and accelerates uniformly at 3 m/s\textsuperscript{2. What will be its velocity after 5 seconds?}

Answer 1

To determine the time t in which the particle moves a distance of 135 meters while its velocity changes from 10 \, ms^{-1} to 20 \, ms^{-1}, we use the equation of motion:

v^2 = u^2 + 2as

where:

v is the final velocity = 20 \, ms^{-1}
u is the initial velocity = 10 \, ms^{-1}
a is the acceleration
s is the distance = 135 m

Substitute the given values into the equation:

(20)^2 = (10)^2 + 2 \cdot a \cdot 135

400 = 100 + 270a

Simplify to find a:

300 = 270a \Rightarrow a = \frac{300}{270} = \frac{10}{9} \, ms^{-2}

Now, use the first equation of motion to find time t:

v = u + at

20 = 10 + \frac{10}{9} \cdot t

Solve for t:

10 = \frac{10}{9} \cdot t

t = \frac{10 \cdot 9}{10} = 9 \, s

Therefore, the time t is 9 seconds.

Answer 2