An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What are (a) its speed at the first point and (b) its acceleration?

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limacarp4 · Accepted Answer

a) Since that is a motion with a constant acceleration, we can use the following two equations vx=vx,0+axt x−xo=vx,ot+12axt2 where (vx=15 ms) is the speed at the second point, (x−xo=70) is the traveled distance and t=6 s. Substitutte for vx,ax and t into the two equations to get the following 15ms=vx,o+(6s)ax rearrange the equation to isolate ax ax=(15 ms)−vx,o6 s 70 m=(6 s)vx,o+(18 s2)ax Substitute for ax from equation (1) into equation (2) 70 m=(6 s)vx,o+(45 m)−(3 s)vx,o rearrange the equation and solve to find vx,o vx,o=(70−45) m3 s=8.33 ms b) The acceleration can be determined by substituting the value of vx,o into equation ax=(15 ms)−8.33 ms6 s=1.11 ms2

karton · Accepted Answer

Answer:(a) The speed of the antelope at the first point is v1≈−8.34 m/s.(b) The acceleration of the antelope is a≈3.89 m/s².Explanation:d = distance between the two points (d=70.0 m)t = time taken to cover the distance (t=6.00 s)v2 = speed at the second point (v2=15.0 m/s)We are asked to find:(a) the speed at the first point (v1)(b) the acceleration (a)We can start by using the formula for displacement with constant acceleration:d=v1t+12at2Since the antelope starts from rest at the first point, its initial speed (v1) is 0. Plugging in the given values, we have:70.0=0·6.00+12a·(6.00)2Simplifying the equation, we get:70.0=18.0aTo find a, we divide both sides of the equation by 18.0:70.018.0=aHence, the acceleration of the antelope is:a≈3.89m/s2For part (a), we can use the formula for velocity with constant acceleration:v2=v1+atWe already know v2 and t, and we just found a. Substituting these values, we can solve for v1:15.0=v1+3.89·6.00Simplifying the equation:15.0=v1+23.34To isolate v1, we subtract 23.34 from both sides:v1=15.0−23.34Hence, the speed of the antelope at the first point is:v1≈−8.34m/sNote: The negative sign indicates that the antelope is moving in the opposite direction at the first point.

user_27qwe · Accepted Answer

Step 1: (a) The speed of the antelope at the first point (v1) can be calculated using the formula for constant acceleration:v2=v1+at where v2 is the speed at the second point, a is the acceleration, and t is the time taken. Rearranging the equation, we have:v1=v2−atSubstituting the given values, we get:v1=15.0m/s−a×6.00sStep 2: (b) The acceleration (a) can be determined using the formula:s=ut+12at2 where s is the distance covered, u is the initial velocity, t is the time, and a is the acceleration. Rearranging the equation, we have:a=2(s−ut)t2Substituting the given values, we get:a=2(70.0m−v1×6.00s)(6.00s)2

alenahelenash · Accepted Answer

To solve the problem, let&#039;s use the following variables:- d for the distance between the two points (given as 70.0 m).- t for the time taken to cover the distance (given as 6.00 s).- v2 for the speed of the antelope at the second point (given as 15.0 m/s).We need to find:(a) The speed of the antelope at the first point (v1).(b) The acceleration of the antelope (a).(a) To find the speed at the first point, we can use the equation for average speed:vavg=dtThe average speed is the total distance divided by the total time. Since the antelope is moving with constant acceleration, the average speed is equal to the average of the initial and final speeds. So we have:vavg=v1+v22Substituting the given values, we get:70.06.00=v1+15.02To find v1, we can rearrange the equation:v1=2(70.06.00)−15.0(b) To find the acceleration, we can use the equation:v2=v1+a·tRearranging the equation to solve for a, we have:a=v2−v1tSubstituting the values we have found, we get:a=15.0−v16.00Now, let&#039;s calculate the values of v1 and a.(a) Calculating v1:v1=2(70.06.00)−15.0(b) Calculating a:a=15.0−v16.00

An antelope moving with constant acceleration covers the distance between

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