Question

URGENT. please help with acceleration equation problems.

Answer 1

(a) See graph in attachment.

The speed decreases from 45 m/s to 25 m/s after 4 seconds, as stated in the text of the problem.

(b) See graph in attachment.

In a distance-time graph, the slope of the graph corresponds to the speed of the object. In this case, we know that the speed is decreasing, so this is why the slope of the curve decreases.

(c) See graph in attachment

The acceleration of the object is constant, because the speed decreases uniformly, so in an acceleration-time graph it corresponds to a flat line. Also, it has a negative value, since it is a deceleration (the speed is decreasing).

(d) -5 m/s^2

The value of the acceleration of the car is given by:

$a=\frac{\Delta v}{t}$

where $\Delta v=v_f-v_i$ is the change in velocity of the car, and t is the time taken. Substituting numbers, we find

$a=\frac{v_f-v_i}{t}=\frac{25 m/s-45 m/s}{4 s}=-5 m/s^2$

(e) $v(t)=45-5t$

The speed at time t of a car in a uniform accelerated motion is given by

$v(t)=v_0 + at$

where v0 is the initial speed, a is the acceleration and t is the time. In this problem, $v_0=45 m/s$ and $a=-5 m/s^2$, so we can substitute the numbers into the equation and we obtain

$v(t)=45 -5t$

(f) 35 m/s

Since the acceleration is constant, the average speed of the car can be calculated by using the following formula:

$v=\frac{v_f+v_i}{2}$

where vf and vi are the final and initial speeds. In this case, $v_f=45 m/s$ and $v_i=25 m/s$, so the average speed is

$v=\frac{45 m/s+25 m/s}{2}=35 m/s$

(g) 140 m

We can calculate the distance traveled by using two different formulas:

1) $S=v_0 t + \frac{1}{2}at^2$

where v0=45 m/s is the initial speed, a=-5 m/s^2 is the acceleration and t=4 s is the time. Substituting, we obtain

$S=(45 m/s)(4 s)+\frac{1}{2}(-5 m/s^2)(4 s)^2=140 m$

2) $S=v_{avg} t$

where $v_{avg}=35 m/s$ is the average speed calculated previously, and t=4 s is the time. Using this formula, we find

$S=(35 m/s)(4 s)=140 m$