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Question: A particle starts from rest accelerates at \(2m{{s}^{-{{2}^{{}}}}}\)for 10s and then goes with const...

A particle starts from rest accelerates at 2ms−22m{{s}^{-{{2}^{{}}}}}for 10s and then goes with constant speed for 30s and then decelerates at 4ms−24m{{s}^{-2}}till it stops after the next. What is the distance travelled by it?
A. 750m
B. 800m
C. 700m
D. 850m

Explanation

Solution

As a first step, one could read the question and hence note down the given values from the question. Then one could recall the equations of motion accordingly. After that, you could simply carry out the necessary substitutions in these equations and hence find the net displacement.

Formula used:
Equations of motion,
s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}}
v2−u2=2as{{v}^{2}}-{{u}^{2}}=2as

Complete step-by-step solution:
In the question, we are given a particle that starts accelerating from rest at 20ms−220m{{s}^{-2}}for 10s and then maintain constant speed for next 30s then again decelerates at 4ms−24m{{s}^{-2}}until it stops after the next. We are supposed to find the distance travelled by it from the above given information.
The displacement for the first 10 seconds would be,
s1=12a1t12=12×2×102=100m{{s}_{1}}=\dfrac{1}{2}{{a}_{1}}{{t}_{1}}^{2}=\dfrac{1}{2}\times 2\times {{10}^{2}}=100m
Now let us find the velocity gained in the first 10s as this is then maintained constant for the next 30s.
v1=u1+a1t1=0+2×10=20ms−1{{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}}=0+2\times 10=20m{{s}^{-1}}
Now the displacement for the next 30 seconds would be,
s2=v1t2=20×30=600m{{s}_{2}}={{v}_{1}}{{t}_{2}}=20\times 30=600m
The displacement during the deceleration would be given by,
s3=v2−v122a=0−2022×(−4)=50m{{s}_{3}}=\dfrac{{{v}^{2}}-{{v}_{1}}^{2}}{2a}=\dfrac{0-{{20}^{2}}}{2\times \left( -4 \right)}=50m
The net displacement for the given particle would be,
S=s1+s2+s3=100+600+50S={{s}_{1}}+{{s}_{2}}+{{s}_{3}}=100+600+50
∴S=750m\therefore S=750m
Therefore, we found the net displacement of the given particle to be 750m.

Note: All we have done in order to solve the given question is mere substitution of the given values into the known equations of motion. All we have to understand is that starting the motion from rest would imply that initial velocity of the body is zero and the uniform velocity would imply that the acceleration is zero.