## Accelerated Motion Common Problems

- Category: High School
- Written by fisikastudycenter

**Fisikastudycenter.com-** The Examples of High School Physics Problems and Solutions for Grade 10, Motion along Straight Line : Constant Velocity, Uniform Accelerated Motion.

**Problem 1**

A stone of 200 g throw up with 50 m/sec of initial velocity.

Ignore the air friction and use the acceleration due to gravity g = 10 m/sec^{2}, find :

a) The maximum height of stone

b) Time taken to reach the maximum height

c) Time taken by the stone before reaching the ground.

**Solution**

a) The stone velocity = 0 m/sec, in it's maximum height. Using accelerated motion formula:

b) The time taken:

c) Time for the stone in the air:

t = (2)(5) = 10 sec

**Problem 2**

A car moves at speed of 72 km/hrs, before the breaks works and finally stops at 8 m after the break works. Find the acceleration of the car!

**Solution**

First, convert the unit of speed from km/hrs to m/sec.Using the formula:

**Problem 3**

Given a graph shown below:

Find:

a. distance for t = 5 sec till t = 10 sec

b. the displacement for t = 5 sec till t = 10 sec

**Solution**

Finding the distance when given a graph of V (speed) - t (time) : Calculate the area of V-t graph (under the lines till t axis), all area is positif value, for calculating the distance, above the t axis also under t axis. For calculating the distance, use the same methode, but the above t axis area is positif, under the t axis is negative value,

**Problem 4**

An ant moves from point A to point B as shown below.

The radius of circle is r = 2m, and the ant need 10 sec to reach point B.Find:

a) The average velocity

b) The average speed

**Solution**

We have to calculate the distance and displacement of the ant first:

Distance = ^{1}/_{4} (2πr) = ^{1}/_{4} (2π x 2) = π m

Displacement, use phytagorean:

Displacement = √ ( 2^{2} + 2^{2} ) = 2√2 meter.

a) Average velocity = displacement : time

Average velocity = 2√2 meter : 10 sec = 0,2√2 m/sec

b) Average speed = distance : time

Average speed = π meter : 10 sec = 0,1 π m/sec

**Problem 5**

An airplane moves from P to Q at the east direction for 30 minutes with constant velocity of 200 km/hrs. From Q the plane moves to R that located 53^{o} from east direction an need 1 hr flaying with constant average of 100 km/hr.

Find:

a) The average velocity of the plane

b) The average speed of the plane

**Solution**

First, find the length of PQ, QR, QR', RR', PR' and PR

PQ = V_{PQ} x t_{PQ} = (200 km/jam) x (0,5) jam = 100 km

QR = V_{QR} x t_{QR} = (100 km/jam) x (1 jam) = 100 km

QR' = QR cos 53^{o} = (100 km) x (0,6) = 60 km

RR' = QR sin 53^{o} = (100 km) x (0,8) = 80 km

PR' = PQ + QR' = 100 + 60 = 160 km

PR = √[ (PR' )^{2} + (RR')^{2} ]

PR = √[ (160 )^{2} + (80)^{2} ] = √(32000) = 80√5 km

the distance = PQ + QR = 100 +100 = 200 km

the displacement = PR = 80√5 km

the time taken = 1 hr + 0,5 hr = 1,5 hrs

a) The average velocity = displacement : time taken = 80√5 km : 1,5 hr = 53,3 √5 km/hr

b) average speed = distance : time taken = 200 km : 1,5 hr = 133,3 km/hr