a. Acceleration is given by
\(a = \frac{{\Delta v}}{{\Delta t}}\)
$\Delta v = 54 -36 = 18 km/hr =\frac {18 \times 1000}{3600} m/s= 5 m/s$
So a=.5 m/s2
b. Distance is given by
\(S = ut + \frac{1}{2}a{t^2}\)
Now u=36 km/hr =10m/s
$s=10 \times 10 + \frac {1}{2} \times .5 \times 10^2$
So s=125m
might be accelerated
Here u=54 km/h =15 m/s ,v=36 km/h=10 m/s
Acceleration is given by
\(a = \frac{{\Delta v}}{{\Delta t}}\)
$a = \frac {10 -15}{10} =- .5$
So a=-.5 m/s2
Negative sign implies retardation
| S.no | Rounds | Displacement | Distance |
|---|---|---|---|
| 1 | 1 | ||
| 2 | 1.5 | ||
| 3 | 2 | ||
| 4 | 2.5 |
After every round, particle comes to it starting position. So displacement at full rounds will be zero
| S.no | Rounds | Displacement | Distance |
|---|---|---|---|
| 1 | 1 | 0 | \(20\pi \) |
| 2 | 1.5 | 20 m | \(30\pi \) |
| 3 | 2 | 0 | \(40\pi \) |
| 4 | 2.5 | 20m | \(50\pi \) |
Acceleration is given by
\(a = \frac{{\Delta v}}{{\Delta t}}\)
$ a = \frac {20 -10}{4} = \frac {10}{4}=2.5$
So a=2.5 m/s2
Here u=0 ,a= 5m/s2 ,t= 5 sec,v=?
Now $ v=u+at$
$v= 0 + 5 \times 5 =25$ m/s
Answer is (c)
$ \text {Average speed}= \frac {\text {total distance}}{time} =\frac {500}{10}=50$ km/hr
Ratio of average speed to maximum speed= 50:90=5:9
u=0, v=54 km/h=15 m/s ,t=2 sec
a) Acceleration is given by
\(a = \frac{{\Delta v}}{{\Delta t}}\)
$a = \frac {15-0}{2}$
So a=7.5 m/s2
b) Distance is given by
\(S = ut + \frac{1}{2}a{t^2}\)
$s= 0 \times 2 + \frac{1}{2} \times 7.5 \times 4$
s=15m
$v=u+at$
\(u = 0\)
\(v = 10 \times 5 = 50m/s\)
As it comes down to the initial point
Net displacement is zero
Net distance=200 m
We know that
$Distance = speed \times time$
Distance travelled in first 2 minute = $\frac {7.5 \times 2}{60}$ = 0.25 km
Distance travelled in last 2 minute = $\frac {7.5 \times 2}{60}$= 0.25 km
Total distance = .25+.25 = 0.5 km
Total time = 2+2+56 = 60 minute = 1 hr
Average speed = 0.5/1
= 0.5 km/hr
$speed =\frac {distance}{time}$
or
$time= \frac {distance}{speed}$
For travelling 1 km with 3 km/hr,
Time taken = 1/3 hr = 20 min
Now he has to reach school in 30 min,So he has cover another 1 km in = 30 -20 =10 min= 1/6 hr
So speed should = distance /time= 1/(1/6) = 6 km/hr
u=10 m/s,t=2 s, a=2 m/s2
$v=u+at$
$v= 10 + 2 \times 2 =14$ m/s
We are given the following information:
Initial velocity, \(u = 20 \, \text{m/s}\)
Distance traveled, \(d = 6 \, \text{cm} = 0.06 \, \text{m}\)
We can use the equation of motion:
\[v^2 = u^2 + 2ad\]
Since the bullet comes to rest (\(v = 0 \, \text{m/s}\)), we have:
\[0 = u^2 + 2ad\]
Solving for deceleration (\(a\)):
\[2ad = -u^2\]
\[a = \frac{{-u^2}}{{2d}}\]
Substituting the given values:
\[a = \frac{{-(20 \, \text{m/s})^2}}{{2 \cdot 0.06 \, \text{m}}}\]
\[a = -\frac{{400 \, \text{m}^2/\text{s}^2}}{{0.12 \, \text{m}}}\]
\[a = -3333.33 \, \text{m/s}^2 \, \text{(rounded to two decimal places)}\]
Therefore, the deceleration of the bullet in the sand box is approximately \(-3333.33 \, \text{m/s}^2\).
(c)
a. The distance travelled by the SUV is first 2 seconds = Area of $\Delta ABE$
$= \frac {1}{2} \times AE \times BE$
$=\frac {1}{2} \times 2 \times 15= 15m$
b. Acceleration will be given by the slope of the line CD
$a= \frac {0 -15}{6-5} = -15 m/s^2$
Now mass of the SUV =1000 Kg
Braking Force will be
$F=ma = 1000 \times -15 =-15000 N$
Given $u= 5 \times 10^3 m/s$ , $a=10^3 m/s^2$, $v=2u =10^4$ m/s
i.Using $v=u+at$
$10^4 = 5 \times 10^3 + 10^3 t$
or
t=5 sec
ii. Using $s=ut+ \frac {1}{2}at^2$
$s = 5 \times 10^3 \times 5 + \frac {1}{2} \times 10^3 \times 5^2$
$=37.5 \times 10^3 m$
Given Lenght of Train = 100m , velocity = 72 km/hr = 20 m/s, Lenght of the Bridge =2 Km.
Total distance covered by the train to fully pass the bridge = 2000 + 100 = 2100 m
So, time taken
$ time = \frac {distance}{velocity} = \frac {2100}{20} =105 sec$
Given r =42.250 km , T= 24 hour
Linear velocity in circular motion is given by
$v = \frac {2 \pi r}{T} = \frac {2 \times 3.14 \times 42.250}{24} = 11.05 km/hr$
(i) We can see from the graph that velocity is not changing So,acceleration is equal to zero.
(ii) By Reading the graph, velocity = 20 m/s
(iii) Distance covered in 15 seconds, $s = u \times t = 20 \times 15 = 300$ m
Here u=0, s=20 m ,t=2 sec
\(S = ut + \frac{1}{2}a{t^2}\)
$20= 0 + \frac {1}{2} a \times 4$
$a=10 m/s^2$
Now Velocity at the end of 2 sec
$v=u+at$
$v=10 \times 2=20$ m/s
Now Checking the motion for next 4 sec, let assume acceleration be $a_1$ as we are not sure about the motion in second part
Here u=20 m/s ,s=160 m ,t=4 sec
\(S = ut + \frac{1}{2}a{t^2}\)
$160 = 20 \times 4 + \frac {1}{2} a_1 \times 16 $
$a_1 = 10 m/s^2$
So acceleration is constant in both the motion.
Now we can easily calculate velocity as
u=0, t= 7sec , a=10m/s2
$v=u+at$
$v= 0 + 10 \times 7= 70 m/s$
Total distance travelled by the object =20 m + 16 m = 36 m
Total time taken = 2 s + 2 s = 4 s
$\text {Average Speed} = \frac {\text {total Distance travelled}}{\text {total time taken }}$
$\text {Average Speed} = \frac {36}{4} = 9$ m/s
(d) only.
Distance and displacement in all other cases
Here u=72 km/hr = $\frac {72 \times 1000}{3600} m/s= 20$ m/s , v=0
a= -5 m/s2
Now using the relation
$v^2 = u^2 + 2as$
$0 = (20)^2 + 2 \times (-5) \times s$
$s= 40 m$
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