Kinematics Question 371
Question: A car, starting from rest, accelerates at the rate through a distance S, then continues at constant speed for time t and then decelerates at the rate $ \frac{f}{2} $ to come to rest. If the total distance traversed it’s 15 S, then
Options:
A) $ \text{S=}\frac{1}{6}f{{t}^{2}} $
B) $ \text{S=ft} $
C) $ \text{S=}\frac{1}{4}f{{t}^{2}} $
D) $ \text{S=}\frac{1}{72}f{{t}^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
distance from A to B = $ S=\frac{1}{2}ft _{1}^{{}} $ distance from B to C = $ ( ft _{1} )t $ distance from C to D = $ \frac{u^{2}}{2a}=\frac{{{( ft _{1} )}^{2}}}{2( f/2 )} $
$ ft _{1}^{2}=2S $
$ \Rightarrow S+ft _{1}t+2S=15S $
$ \Rightarrow ft _{1}t+12S\text{ }…..\text{(i)} $
$ \frac{1}{2}ft _{1}^{2}=S $ Dividing (i) by (ii), we get $ t _{1}=\frac{t}{6} $
$ \Rightarrow S=\frac{1}{2}f{{( \frac{t}{6} )}^{2}}=\frac{ft^{2}}{72} $
BETA
