The displacement is given by x = 22 +t+5, the acceleration at t = 5 sec will be
8 m/s2
O 12 m/s2
O 15 m/s2
O
4 m/s2
Answers
Answer:
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.
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 m/s2.
If a particle has an initial velocity of 6 m/s and a constant acceleration of −2 m/s2, then:
when t=1, the velocity of the particle is 4 m/s
when t=2, the velocity of the particle is 2 m/s
when t=3, the velocity of the particle is 0 m/s
when t=4, the velocity of the particle is −2 m/s
when t=10, the velocity of the particle is −14 m/s.
In general, the velocity of the particle is 6−2t m/s after t seconds. The velocity–time graph for this motion is shown below; it is the graph of v(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.
More precisely, the constant acceleration a is given by the formula
a=v(t2)−v(t1)t2−t1,
where v(ti) is the velocity at time ti. Since velocity is a vector, so is acceleration.
Example
A particle is moving in a straight line with constant acceleration of 1.5 m/s2. Initially its velocity is 4.5 m/s. Find the velocity of the particle:
after 1 second
after 3 seconds
after t seconds.
Solution
After 1 second, the velocity is 4.5+1.5=6 m/s.
After 3 seconds, the velocity is 4.5+3×1.5=9 m/s.
After t seconds, the velocity is 4.5+1.5t m/s.
Example
A car is travelling at 100 km/h =2509 m/s, and applies its brakes to stop. The acceleration is −10 m/s2. How long does it take for the car to stop?
Solution
After one second, the car's velocity is 2509−10 m/s. After t seconds, its velocity is
v(t)=2509−10t m/s.
The car stops when v(t)=0. Solving this equation gives
2509−10tt=0=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.)
The constant-acceleration formulas for motion in a straight line
Throughout this section, we have been considering motion in a straight line with constant acceleration. This situation is very common; for example, a body moving under the influence of gravity travels with a constant acceleration.
There are five frequently used formulas for motion in a straight line with constant acceleration. The formulas are given in terms of the initial velocity u, the final velocity v, the displacement (position) x, the acceleration a and the time elapsed t. Of course, they require consistent systems of units to be used.
It is assumed that the motion begins when t=0, and that the initial position is taken as the origin, that is, x(0)=0.
The five equations of motion
v=u+at
x=(u+v)t2
x=ut+12at2
v2=u2+2ax
x=vt−12at2
Note. Each of the five equations involve four of the five variables u,v,x,a,t. If the values of three of the variables are known, then the remaining values can be found by using two of the equations.
Deriving the constant-acceleration formulas
The first equation of motion