(12 m s-1 ) x (2 s) +1/2
(-6 m s-2) (2 s)
Answers
Step-by-step explanation:
Let t be the time in seconds from the beginning of the motion of a particle. If the particle has a velocity of 4 m/s initially (at t=0) and has a constant acceleration of 2 m/s2, find the velocity of the particle:
when t=1
when t=2
after t seconds.
Draw the velocity–time graph for the motion.
Solution
When t=1, the velocity of the particle is 4+2=6 m/s.
When t=2, the velocity of the particle is 4+2×2=8 m/s.
After t seconds, the velocity of the particle is 4+2t m/s.
Step-by-step explanation:
Constant acceleration
We are all familiar with the fact that a car speeds up when we put our foot down on the accelerator. The rate of change of the velocity of a particle with respect to time is called its acceleration. If the velocity of the particle changes at a constant rate, then this rate is called the constant acceleration.
Since we are using metres and seconds as our basic units, we will measure acceleration in metres per second per second. This will be abbreviated as m/s22. It is also commonly abbreviated as ms−2−2.
For example, if the velocity of a particle moving in a straight line changes uniformly (at a constant rate of change) from 2 m/s to 5 m/s over one second, then its constant acceleration is 3 m/s22.
Example
Let tt be the time in seconds from the beginning of the motion of a particle. If the particle has a velocity of 4 m/s initially (at t=0t=0) and has a constant acceleration of 2 m/s22, find the velocity of the particle:
when t=1t=1
when t=2t=2
after tt seconds.
Draw the velocity–time graph for the motion.
Solution
When t=1t=1, the velocity of the particle is 4+2=64+2=6 m/s.
When t=2t=2, the velocity of the particle is 4+2×2=84+2×2=8 m/s.
After tt seconds, the velocity of the particle is 4+2t4+2t m/s.
Decreasing velocity
If the velocity of a particle moving in a straight line changes uniformly (at a constant rate of change) from 5 m/s to 2 m/s over one second, its constant acceleration is −3−3 m/s22.
If a particle has an initial velocity of 6 m/s and a constant acceleration of −2−2 m/s22, then:
when t=1t=1, the velocity of the particle is 4 m/s
when t=2t=2, the velocity of the particle is 2 m/s
when t=3t=3, the velocity of the particle is 0 m/s
when t=4t=4, the velocity of the particle is −2−2 m/s
when t=10t=10, the velocity of the particle is −14−14 m/s.
In general, the velocity of the particle is 6−2t6−2t m/s after tt seconds. The velocity–time graph for this motion is shown below; it is the graph of v(t)=6−2tv(t)=6−2t.
Over the first three seconds, the particle's speed is decreasing (the particle is slowing down). At three seconds, the particle is momentarily at rest. After three seconds, the velocity is still decreasing, but the speed is increasing (the particle is going faster and faster).
Summary
If we assume that the rate of change of velocity (acceleration) is a constant, then the constant acceleration is given by
Acceleration=Change in velocityChange in time.
Acceleration=Change in velocityChange in time.
More precisely, the constant acceleration aa is given by the formula
a=v(t2)−v(t1)t2−t1,
a=v(t2)−v(t1)t2−t1,
where v(ti)v(ti) is the velocity at time titi. Since velocity is a vector, so is acceleration.
Example
A particle is moving in a straight line with constant acceleration of 1.5 m/s22. Initially its velocity is 4.5 m/s. Find the velocity of the particle:
after 1 second
after 3 seconds
after tt seconds.
Solution
After 1 second, the velocity is 4.5+1.5=64.5+1.5=6 m/s.
After 3 seconds, the velocity is 4.5+3×1.5=94.5+3×1.5=9 m/s.
After tt seconds, the velocity is 4.5+1.5t4.5+1.5t m/s.
Example
A car is travelling at 100 km/h =2509=2509 m/s, and applies its brakes to stop. The acceleration is −10−10 m/s22. How long does it take for the car to stop?
Solution
After one second, the car's velocity is 2509−102509−10 m/s. After tt seconds, its velocity is
v(t)=2509−10t m/s.
v(t)=2509−10t m/s.
The car stops when v(t)=0v(t)=0. Solving this equation gives
2509−10tt=0=259.
2509−10t=0t=259.
The car takes approximately 2.8 seconds to stop.
(In exercise 6, we will find out how far the car travels during this time