Question

A particle having initial velocity 5m/s moves with a constant acceleration 2m/s^2 for a time 10seconds along a straight line. Find the displacement of particle in the last one second and the total distance traveled in 10 seconds.



no spamming.....​

Answer 1

Answer:

150m

Explanation:

s=ut+1/2at*t

5*10+1/2*2*10*10

=150m

Answer 2

Given :

▪ Initial velocity = 5mps

▪ Acceleration = 2m/s$^2$

▪ Time interval = 10s

To Find :

▪ Displacement of particle in the last second.

▪ Total distance covered by particle in the given interval of time.

Concept :

➡ Since, acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this type of questions.

Calculation :

⚾ Displacement in last second :

$\implies\bf\:d_{n^{th}}=u+\dfrac{a}{2}(2n-1)\\ \\ \implies\sf\:d_{10^{th}}=5+\dfrac{2}{2}(2×10-1)\\ \\ \implies\sf\:d_{10^{th}}=5+1(19)\\ \\ \implies\underline{\boxed{\bf{\red{d_{10^{th}}=24m}}}}$

⚾ Total distance travelled in 10s :

$\Rightarrow\bf\:d=ut+\dfrac{1}{2}at^2\\ \\ \Rightarrow\sf\:d=(5\times 10)+\dfrac{1}{2}(2\times10^2)\\ \\ \Rightarrow\sf\:d=50+\dfrac{200}{2}\\ \\ \Rightarrow\sf\:d=50+100\\ \\ \Rightarrow\underline{\boxed{\bf{\blue{d=150m}}}}$